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WELLS'    MATHEMATICAL   SERIES. 


Academic  Arithmetic. 

Academic  Algebra. 

Higher  Algebra. 

University  Algebra. 

College  Algebra. 

Plane  Geometry. 

Solid  Geometry. 

Plane  and  Solid  Geometry. 

Plane  and  Spherical  Trigonometry. 

Plane  Trigonometry. 

Essentials  of  Trigonometry. 

Logarithms  (flexible  covers). 

Elementary  Treatise  on  Logarithms. 


Special  Catalogue  and  Terms  on  application. 


^- 


THE 


ESSENTIALS 


OF 


PLAXE    AXD    SniERICAL 
TRIGOXOMETllY. 


BY 

WEBSTER   WELLS,   S.B., 

Associate  Professor  of  Mathematics  in  the  Massachusetts 
Institute  of  Technology. 


coU.»o»,-^' 


•iSr^^^  ^'""^ 


irUkTrt 


.  oerT' 


LEACH,   SHEWELL,   AND   SANBOEN. 
BOSTON.     NEW  YORK.     CHICAGO. 


Copyright,  1887, 
By  WEBSTER  \YELLS, 


BOSTON  COLLEGE  LIBRARY 
CHESTNUT  HILL,  MASS. 


Typography  by  Presswork  by 

J.  S.  Gushing  &  Co.  Berwick  &  Smith. 


PEEFACE. 


This  volume  contains  only  those  portions  of  Plane  and 
Spherical  Trigonometry  which  are  essential  in  the  practical 
applications  of  the  subject  to  problems  in  surveying,  geod- 
esy, and  navigation. 

The  attention  of  teachers  and  others  is  invited  to  the  fol- 
lowing, which  may  be  regarded  as  constituting  the  salient 
features  of  the  work : 

1 .  The  proofs  of  the  functions  of  0°,  90°,  180°,  and  270°  ; 
Arts.  38  to  41. 

2.  The  figures,  and  the  tabular  arrangement  of  the  work, 
in  the  discussions  of  Arts.  42  and  44. 

3.  The  solution  of  problems  in  Art.  56  by  the  construc- 
tion of  a  diagram,  in  a  manner  analogous  to  that  of  Art.  15. 

4.  The  proofs  of  the  fundamental  formuhe  for  any 
angle  ;    Arts.  57  to  59. 

5.  The  discussion  of  the  line  values  of  the  trigonometric 
functions,  and  their  application  in  tracing  the  changes  in 
the  functions  as  the  angle  increases  from  0°  to  360°  ;  Arts. 
62  to  64. 

6.  The  proofs  of  the  formulje  for  the  sines  and  cosines 
of  X  -\-  y  and  x  —  y  for  an}'  values  of  x  and  y ;  Arts.  68 
to  70. 

7.  The  discussion  of  the  ambiguous  case  in  the  solution 
of  plane  oblique  triangles  ;  Arts.  124  to  126. 


iy  PREFACE. 

8.  The  geometrical  proofs  (Art.  138)  of  the  propositions 
that  in  any  spherical  riglit  triangle  : 

I.  If  the  sides  including  the  right  angle  are  in  the  same 
quadrant,  the  hypotenuse  is  <  90° ;  if  they  are  in  different 
quadrants,  the  hypotenuse  is  >  90°. 

II.    An  angle  is  in  the  same  quadrant  as  its  opposite  side. 

9.  The  discussion  of  the  properties  of  spherical  right 
triangles  before  those  of  spherical  oblique  triangles ;  see 
Chapters  XI.  and  XII. 

10.  The  reduction  of  the  number  of  cases  in  the  complete 
demonstration  of  the  fundamental  formulae  for  spherical 
right  triangles,  to  three,  by  application  of  the  results  proved 
geometrically  in  Art.  138  ;  see  Art.  143. 

11.  The  discussions  of  the  ambiguous  cases  in  the  solution 
of  spherical  oblique  triangles  (Arts,  171  and  172)  ;  espe- 
cially the  rules  given  on  pages  130  and  132  for  determining 
the  number  of  solutions. 

At  the  end  of  the  book  will  be  found  a  collection  of  for- 
mulae in  form  for  convenient  reference. 

Teachers  who  desire  a  briefer  course  are  recommended  to 
omit  Chapter  IV.,  which  may  be  done  without  interrupting 
the  logical  completeness  of  the  rest  of  the  work.  Chapter 
VI.  may  also  be  omitted  by  those  who  have  taken  the  sub- 
ject of  Logarithms  in  their  course  in  Algebra.  The  course 
might  be  still  further  abridged,  if  desired,  by  the  omission  of 
the  exercises  at  the  end  of  Chapter  V. 

WEBSTER  WELLS. 
Boston,  1887. 


CONTENTS. 


PART  I.      PLANE   TRIGONOMETRY. 

Page 

T.     Definitions  ;   Measurement  of  Angles  ....  1 

Circular  Measure  of  an  Angle 1 

IT,     The  Trigonometric  Functions 4 

Functions  of  Acute  Angles 4 

Fundamental  Theorems 10 

III.  Application  of  Algebraic  Signs 12 

Trigonometric  Functions  of  Angles  in  General   .     .  12 

Rectangular  Co-ordinates 14 

General  IVfinitions  of  the  Functions 15 

Functions  of  0°  90°  180°  270°,  and  360°  ....  19 

Functions  of  (—yl)  in  Terms  of  those  of  ^     ...  21 

Functions  of  (90°  +  .4)  in  Terms  of  those  of  ^1   .     .  24 

Proofs  of  the  Fundamental  Formuhu  for  any  Angle  34 

IV.  Miscellaneous  Theorems 36 

To  express  Each  of  the  Six  Principal  Functions  in 

Terms  of  the  other  Five 36 

T  •    ••  •      -IT  1          (•  sin X       .  tan  x  oir 

Limiting  Values  of and 37 

XX 

Line  Values  of  the  Trigonometric  Functions  ...  38 

V.    General  Formula 42 

Functions  of  2  a: 48 

Functions  of  ^  a; 49 

Inverse  Trigonometric  Functions 50 

VI.    Logarithms 54 

Properties  of  Logarithms 56 

Applications 61 

Arithmetical  Complement 63 

VII.     Solution  of  Right  Triangles 67 

Formulae  for  the  Area  of  a  Right  Triangle ....  73 


yi  CONTENTS. 

Page 

VIIL     General  Properties  of  Triangles 75 

Formulae  for  the  Area  of  an  Oblique  Triangle     .     .       81 

IX.     Solution  of  Oblique  Triangles 83 

Area  of  an  Oblique  Triangle 92 


PART  11.      SPHERICAL   TRIGONOMETRY. 

X.     Geometrical  Definitions  and  Principles  ...  95 

XI.     Spherical  Right  Triangles 99 

Napier's  Rules  of  Circular  Parts 104 

Solution  of  Spherical  Right  Triangles 106 

Solution  of  Quadrantal  Triangles 112 

XII.     Spherical  Oblique  Triangles .  114 

General  Properties  of  Spherical  Triangles  ...  114 

Napier's  Analogies '     .  121 

Solution  of  Spherical  Oblique  Triangles     ....  123 

Applications 133 


Formula .    137 

Answers  to  the  Examples 144 


Paet   L 
PLAIs^E  TRIGOXOMETRY. 


JXKc 


I.   DEFINITIONS;   MEASUREMENT 
OF  ANGLES. 

1.  Trigonometi^  is  that  branch  of  mathematics  in  which 
algebraic  processes  are  used  to  treat  of  the  properties  and 
measurement  of  geometrical  figures. 

In  Plane  Trigonometry  we  consider  j^Zajie  figures  only. 

2.  An  angle  is  measured  by  finding  its  ratio  to  anothei 
angle  adopted  ar])itrarily  as  the  unit  of  measurement. 

3.  The  usual  unit  of  measurement  for  angles  is  the  degree, 
or  an  angle  equal  to  the  ninetieth  part  of  a  right  angle. 

To  express  fractional  parts  of  the  unit,  the  degree  is  di- 
vided into  sixty  equal  parts,  called  minutes,  and  the  minute 
into  sixty  equal  parts,  called  seconds. 

Degrees,  minutes,  and  seconds  are  denoted  by  the  sym- 
bols °,  ',  ",  respectively  ;  thus,  43°  22'  37"  denotes  an  angle 
of  43  degrees,  22  minutes,  and  37  seconds. 

CIRCULAR    MEASURE    OF    AN    ANGLE. 

4.  Another  method  of  measuring  angles,  and  one  of  great 
importance,  is  known  as  tlie  circular  method,  'in  which  the 
unit  of  measurement  is  the  angle  subtended  at  the  centre  of  a 
circle  by  an  arc  ivhose  length  is  equal  to  the  radius  of  the  circle. 


^  PLANE   TRIGONOMETRY. 

5.    Let  AOB  be  any  angle,  and  AOO  the  unit  of  circular 

measure. 

C 
B 


By  Geometry,  we  have 

angle  ^O-g  _  arc  AB 
angle  ^0(7      slycAG 

arc  AB 
Whence,   circular  measure  ^05  =  — ;-— —  (Art.  4). 

OA 

That  is,  the  circular  measure  of  an  angle  is  equal  to  the  ratio 
of  its  subtending  arc  to  the  radius  of  the  circle. 

For  example,  the  circular  measure  of  a  right  angle  is  equal 
to  the  ratio  of  one-fourth  the  circumference  to  the  radius. 

But  the  circumference  of  a  circle  is  equal  to  the  radius 
multiplied  by  27r,  where  tt  =  3.14159265  ... 

Hence,  if  R  denotes  the  radius,  we  have 


cu'cular  measure  of  90°  = 


i0f27ri?  TT 


B 


6.    Since  the  circular  measure   of  90°  is  -,  the  circular 

measure  of  180°  is  tt  ;  of  60°,  -  ;  of  45°,  -  ;  of  30°,  ^  ;  etc. 

3  4  6 

That  is,  an  angle  expressed  in  degrees  may  be  reduced  to 
circular  measure  by  finding  its  ratio  to  180°,  and  multiplying 
the  result  by  tt. 

23 
Thus,  sinfte  115°  is  —  of  180°,  the  circular  measure  of 

36 

115°  is  ?^. 
36 


DEFIXITI0:N^S  ;    MEASUREMENT   OF   AXGLES.        3 

7.  Conversely,  any  angle  expressed  in  circular  measure 
may  he  reduced  to  degrees  by  multiplying  by  180°  and  dividing 
by  IT ;  or,  more  briefly,  by  substituting  180° /or  tt. 

Thus,  l!I  =  -I  of  180°  =  84°. 
15      15 

8.  In  the  circular  method  such  expressions  may  occur  as 

"  the  angle  -,"  "  the  angle  1,"  etc. 
3 

These  refer  to  the  unit  of  circular  measure  ;  that  is,  the 

2 
angle  -  means  an  angle  whose  subtending  arc  is  two-thirda 
3 

of  the  radius. 

The  angle  1,  that  is,  the  angle  whose  subtending  arc  is 
equal  to  the  radius,  or  the  unit  of  circular  measure,  if 
reduced  to  degrees  l)y  the  rule  of  Art.  7,  gives 

180  180  _„  ^^^_^,^o 

TT         3.14150265... 

The  rule  of  Art.  7  may  then  be  modified  as  follows : 

Any  angle  expressed  in  circular  measure  may  be  reduced  tc 
degrees  by  multiplying  by  57.2957795°... 

Thus,  the  angle  ?  =  -  x  57.2957795°... 
^33 

=  38.1971863°...  =  38°  11' 49.87068"... 


EXAMPLES. 
9.    Express  the  following  in  circular  measure  : 

1.  135°.        3.  11°  15'.        5.  29°  15'.  7.  128°  34f. 

2.  198°.        4.  37°  30'.        6.  174°  22' 30".  8.  92°  48' 45". 

Express  the  following  in  degree  measure  : 

9.  1            11.  11^.        13.  ?.  15.  ^^-^ 

2                        30                  4  3 

10.  — .         12.  — .           14.  2.  16.  ^~^. 

5                     4  4 


PLANE   TRIGONOMETRY. 


II.    THE  TRIGONOMETRIC  FUNCTIONS. 

FUNCTIONS   OF  ACUTE  ANGLES. 
10.    Let  BAG  be  any  acute  angle. 


From  any  point  in  either  side,  as  B,  draw  BO  perpendic- 
ular to  the  other  side,  forming  a  right  triangle  ABC. 

We  then  have  the  following  definitions,  applicable  to  either 
of  the  acute  angles  A  and  B : 

In  any  right  triangle^ 

The  SINE  of  either  acute  angle  is  the  ratio  of  the  op- 

posite side  to  the  hypotenuse. 

The  COSINE  is  the  ratio  of  the  adjacent  side  to  the  hypot- 
enuse. 

The  TANGENT  is  the  ratio  of  the  opposite  side  to  the  adja- 
cent side. 

The  COTANGENT  is  the  ratio  of  the  adjacent  side  to  the  oppo- 
site side. 

The  SECANT  is  the  ratio  of  the  hypotenuse  to  the  adjacent 
side. 

The  COSECANT  is  the  ratio  of  the  hypotenuse  to  the  opposite 
side. 

That  is,  denoting  the  sides  BC,  CA,  and  AB  by  a,  b,  and 
c,  and  employing  the  usual  abbreviations, 


.      a 

sm  A  =  -t 

c 


cos  A  =  -'i 
c 


a 


tan  A  =  -') 


cot  A=-') 
a 


sec  ^  =  -5 
b 


csc^  =  -• 
a 


(1) 


THE   TRIGONOMETRIC   FUNCTIONS. 


And  in  like  manner, 


sin  B  =  1 

c 

COS  B  =  -i 

c 


tan  B  =  -1 

a 


cot  B  =  —I 


0    ^ 
seeB  =  -->   I 
a 


V 


-n        ^ 
CSC  B=  -' 

b 


(2) 


11.    The  following  definitions  are  also  used : 

The  versed  sine  of  an  angle  is  equal  to  unity  minus  the 
cosine  of  the  angle. 

TJie  coversed  sine  is  equal  to  unity  7ninus  the  sine. 

That  is,  vers  A=  1  — ?      covers  A=  1 » 

c  c 


vers  B=l  — 


a 


covers  B=  1 


12.  The  eight  ratios  defined  in  Arts.  10  and  11  are  called 
Trigonometric  Functions,  or  Trigonometric  Jiatios,  of  the 
angle. 

It  is  important  to  observe  that  their  values  depend  solely 
on  the  magnitude  of  the  angle,  and  are  entirely  independent 
of  the  lengths  of  the  sides  of  the  right  triangle  which  con- 
tains it. 


f  yD 


c    c 


"^T—E 


Thus,  let  B  and  B'  be  any  two  points  in  the  side  AD  of 
the  angle  DAE,  and  di-aw  BC  and  B'C  perpendicular  to  AJE. 
Then  by  Art.  10,  we  have 

.      .      BC        ,    .     ,      B'C 
sm  A  =  — — ,  and  sm  A  =    ^  ^,  ♦ 
AB  AB' 


g  PLANE   TRIGONOMETRY. 

But  the  right  triangles  ABO  and  AB'C^  are  similar,  since 
they  have  the  angle  A  common  ;  and  therefore,  by  Geometry, 

'  BC  ^B'C 

AB      AB'' 

Thus  the  two  values  obtained  for  sin  A  are  seen  to  be  equal. 

13.  We  obtain  from  (i)  and  (2),  Art.  10, 

a  =  c  sin  ^,  6  =  c  sin  jB, 

b  =  c  cos  A,  a  =  c  cos  B. 

That  is,  in  any  right  triangle,  either  side  about  the  right 
angle  is  equal  to  the  hypotenuse  multiplied  by  the  sine  of  the 
opposite  angle,  or  by  the  cosine  of  the  adjacent  angle. 

14.  We  have  from  Arts.  10  and  11, 

a  h 

sin  ^  =  -  =  cos  B,  smB  =  -  =  cos  A, 

c  c 

a  b 

tan  A  =  T  =  cot  J5,  tan  B  =  -  =  cot  A, 

b  a 

c  c 

seGA  =  T  =  CSC  B,  secB=-  =  esc  A, 

b  a 

b                                                             a 
vers  A=l  —  =  covers  B,  vers  B=  1 =  covers  A. 

c  c 

Since  the  angles  A  and  B  are  complements  of  each  other, 
the  above  results  may  be  expressed  as  follows : 

The  sine,  tangent,  secant,  and  versed  sine  of  an  acute  angle 
are  respectively  the  cosine,  cotangent,  cosecant,  and  coversed 
sine  of  the  complement  of  the  angle. 

It  is  from  this  circumstance  that  the  names  co-sine,  co- 
tangent, etc.,  were  derived. 

15.  The  Pythagorean  Theorem  affords  a  simple  method 
for  finding  the  values  of  the  remaining  seven  functions  of  an 
acute  angle,  when  the  value  of  any  one  is  given. 


The  trigoxometric  functions. 


1.    Given  cot  ^=  2  ;  requii*ed  the  values  of  the  remaining 
functions  of  A. 

B 


The  equation  may  be  written  cot  A  =  — 

Then  since  the  cotangent  is  the  ratio  of  the  adjacent  side 
to  the  opposite  side,  we  may  regard  our  vahie  as  having  been 
taken  from  a  right  triangle  ABC\  having  its  side  AC  adjacent 
to  the  angle  A  equal  to  2  units,  and  its  side  BC  opposite  to 
A  equal  to  1  unit. 

But  by  Geometry,  we  have 


AB  =  V^.lC'-f  BC  =  V^4  H-  1  =  V^S. 
Whence  bv  definition, 


sin  A  =  — 


1   _Vo 


cos  J. 


sec  A  =  ^^—^ 
2 


tan  A  =  -1 
2 


esc 


vers 


A=s/5, 

A=l-'^. 
5 


A       1        V^ 
covers  A  =  l  ~  j— . 

5 


2.    Given  covers  A  =  -;  required  the  values  of  the  remain- 
ins  functions  of  A. 


PLANE   TKIGONOMETRY. 


2      3 

Since  covers  A=l  —  sin  A^  we  have  sin  A=  1 =  -• 

5      5 

We  then  take  the  opposite  side  BC  equal  to  3  units,  and 
the  hypotenuse  AB  equal  to  5  units. 


Then,  AC==^AB'  -  BO'^  =^25 -9  =  4. 


Whence  by  definition, 

4 

cos  A  =  -•) 
5 


tan  A  =  --) 

4 


cot  A  —  —> 
3 

sec  A  =  -i 

4 


CSC  A  =  -f 
3 

.      1 

vers  A  =  -- 
5 


EXAM  PLES. 
In  each  case  find  the  values  of  the  remaininor  functions  : 


3.    tan^  =  -. 
3 


6.    vers  A  =  — 
4 

X 


9.    cotA  =  x. 


4.    covers  ^  =  —       7.    sin  J.    =-•       10.    cos^ 


y 


17 


5.    CSC  ^  =  4. 


8.    sec^    = —    11.    sec^=-'^ — 

5  b 


16.    To  find  the  values  of  the  sine^  cosine,  tangent,  cotan^^ 
gent,  secant,  and  cosecant  of  45°. 


Let  ABC  be  an  isosceles  right  triangle,  having  each  of  the 
sides  AC  and  BC  equal  to  1. 


Then,        AB  =  s/aC'  +  BC'  =  \/l  +  1  =  ^2. 

Also,  the  angles  A  and  B  are  equal ;  and  since  their  sum 
is  a  right  angle,  each  is  equal  to  45°. 


THE   TRIGOXOMETRIC   FUNCTIONS. 


Then  by  definition, 


sin  45°  =  -4  =  ^.        tan  45°  =  1 .        sec  45°  =  sJ'2. 

\/2       2 


\/2        2 


cot  45°  =  1 


CSC 


45°  =  \/2. 


The  second  line  might  have  been  derived  from  the  first  by 
aid  of  Art.  14,  since  45°  is  complement  of  itself. 

17.     To  Jind  the  values  of  the  sine,  cosine,  tangent,  cotan- 
gent, secant,  and  cosecant  of  30°  and  60°« 


Let  ABD  be  an  equilateral  triangle,  having  each  side  equal 
lo  2.     Draw  AC  perpendicular  to  BD  ;  then  by  Geometr3', 

BC=IBD=\,  and  Z BAC  =l-Z BAD  =  30°. 


Again,     AC  =  ^ Alf  -  BC""  =  ^4  -  1  =  y/3. 
Then  from  the  triangle  xiBC,  by  definition. 


sin  30°  = 


=  cos  60°. 


tan30°  =  i=^   =  cot  60°. 
V3        o 


o_  2  _2V^3_ 


sec  30°  =  -4=  = 


v/^ 


=  CSC  60  . 


cos  30°  =  ^  =  sin  60°. 
2 

cot30°  =  v/3=tan60°. 


CSC  30°  =  2     =  sec  60°. 


Or  the  functions  of  60°  may  be  derived  from  those-'of  30^^ 
by  aid  of  Art.  14,  since  60°  is  the  complement  of  30°, 


10 


PLANE   TRIGONOMETRY. 


FUNDAMENTAL  THEOREMS. 


18.   We  obtain  from  the  definitions  of  Art.  10, 


Sin  A  CSC  ^  =  -  X  -  =  1 . 
c      a 


A  A  b  C  1 

COS  A  sec  A  =  -  x  -  =  1 . 

c       b 

tan  A  cotA  =  -x  -=1. 


These  results  may  be  written 


sin  A  = 
cos  A  = 


1 


CSC  A 

1 

sec  A 


tan^  = 
cot  A  = 


cot  ^' 

1 
tan^ 


sec  A  — 


cos  A 


^ 


}■        (3^ 


CSC  A  = 


sin  A 


That  is,  tJie  sine  of  an  acute  angle  is  the  reciprocal  of  the 
cosecant^  the  tangent  is  the  reciprocal  of  the  cotangent^  and  the 
secant  is  the  reciprocal  of  the  cosine. 

19.    To  j)rove  the  formula 

sin-  A  +  cos-  ^4=1. 

Note.    Sin^^  signifies  (sin  A)'^;   that  is,  the  square  of  the  sine  of  A. 

By  Geometry,  cr  +  6^  =  c^. 

Dividing  by  c^,  gY  +  ('^Y=l. 


THE   TRIGONOMETRIC   FUNCTIONS.  H 

Whence  by  definition, 

{sin  Ay -\-{qos  Ay- =  1; 
or,  sin^  A  +  cos-  A=  1 .  (4) 

The  result  may  be  written  in  the  forms 


sin 


^1  =  V 1  —  cos'-  A,  acd  cos  A  =  y/l  —  sin-  A. 


20.    To  prove  the  formulcB 


,        .      sin  A         1      .    A      cos  A 

tan  A  = ,  and  cot  A  = 

cos  A  sin  A 


a 


Bv  definition,    tan  .1  =  -  =  ^  =  ^HL4,  (5) 


c 


-,                                4.    <       ^       '"       cos  ^  ,  X 

and  cot  ^-1  =  -  =  -  = (g) 

a      t<      sin  J. 
c 

21.    7'o  prove  /7ie  formulce 

sec-  .l  =  1  4-  tan-  A,  and  csc^  ^=  1  +  cot"  A. 

By  G  eom etry ,  c-  =  cr  -f  b- . 

Dividing  by  5-,  ^  =  1  +  ^. 

That  is,  sec-  .4  =  14-  tan-'  A.  (7) 

Dividing  by  a%  -  =  1  +  -.• 

a-  a- 

That  is,  csc^^l  =  1  +  cot-  A.  (s) 


12 


PLANE   TKIGONOMETRY. 


III.    APPLICATION   OF  ALGEBRAIC  SIGNS. 

TRIGONOMETRIC    FUNCTIONS    OF   ANGLES 
IN   GENERAL. 

22.  In  Geometry  we  are,  as  a  rule,  concerned  with  angles 
less  than  two  right  angles  ;  but  in  Trigonometry  it  is  con- 
venient to  regard  them  as  unrestricted  in  magnitude. 


A"' 


In  the  circle  AA",  let  AA"  and  A' A'"  be  a  pair  of  perpen- 
dicular diameters. 

Suppose  a  radius  OB  to  start  from  the  position  OA,  and 
revolve  about  the  point  0  as  a  pivot  in  the  direction  of  OA'. 

When  it  coincides  with  OA',  it  has  described  an  angular 
magnitude  of  90° ;  when  it  coincides  with  OA",  of  180°  ; 
with  OA'",  of  270° ;  with  OA,  its  starting-point,  of  360° ; 
with  OA'  again,  of  450°  ;  etc.  We  thus  see  that  a  signifi- 
cance may  be  attached  to  any  positive  angular  magnitude. 

23.  The  interpretation  of  an  angle  as  the  measure  of  the 
amount  of  rotation  of  a  moving  radius,  enables  us  to  distin- 
guish between  positive  and  negative  angles. 

Thus,  if  a  x)ositive  angle  is  taken  to  indicate  revolution 
from  the  position  OA  in  the  direction  of  OA',  a  negative 
angle  would  naturally  be  taken  as  signifying  revolution  from 
the  position  OA'  in  the  opposite  direction,  or  towards  OA'". 


APPLICATIOX   OF   ALGEBRAIC    SIGNS.  13 

That  is,  if  the  angles  AOB  and  AOB'  are  each  equal  to 
30°  in  absolute  value,  we  should  sa}-  that  AOB  =  +30°,  and 
AOB'  =  -S(f. 

We  may  thus  conceive  of  negative  angles  of  any  magni- 
tude whatever.  It  is  immaterial  which  du'ection  we  consider 
the  positive  direction  of  rotation ;  but  having  adopted  a 
certain  direction  as  positive,  our  subsequent  operations  must 
be. in  accordance. 

24.  The  fixed  line  OA,  from  which  the  rotation  is  sup- 
posed to  commence,  is  called  the  initial  line,  and  the  final 
position  of  the  rotating  radius  is  called  the  terminal  line. 

Either  side  of  an  angle  may  be  taken  as  the  initial  line, 
the  other  being  then  the  terminal  line  ;  thus,  in  the  angle 
AOB,  we  may  consider  OA  the  initial  line  and  OB  the 
terminal  line,  in  which  case  the  angle  is  positive  ;  or  we  may 
consider  OB  the  initial  line  and  OA  the  terminal  line,  in 
which  case  the  anoxic  is  negative. 

25.  In  designating  an  angle,  we  shall  always  write  first 
the  letter  at  the  extremity  of  the  initial  line  ;  thus,  in  desig- 
nating the  angle  AOB,  if  we  regard  OA  as  the  initial  line, 
we  should  call  it  AOB,  and  if  we  regard  OB  as  the  initial 
line,  we  should  call  it  BOA. 

26.  There  are  always  two  angles  in  absolute  value  less 
than  3G0°,  one  positive  and  the  other  negative,  formed  by  a 
given  initial  and  terminal  line. 

Thus,  there  are  formed  by  OA  and  OB'  the  positive  angle 
AOB'  greater  than  270°,  and  the  negative  angle  AOB'  less 
than  90°. 

We  shall  distinguish  between  such  angles  by  referring  to 
them  as  "the  positive  angle  AOB',''  and  "the  negative 
angle  AOB',"  respectively. 

27.  It  is  evident  that  the  terminal  lines  of  two  angles 
which  differ  by  a  multiple  of  360°  are  coincident ;  thus,  the 
angles  30°,  390°,  —330°,  etc.,  have  the  same  terminal  line. 


14 


PLANE   TRIGONOMETRY. 


RECTANGULAR   CO-ORDINATES. 
Y 


a 

h 


N 
a 

-P3  i-bra) 


P,{h,a) 


a 


0  M 

a 


-X 


PiihrO) 


28.  Let  XX'  and  YY'  be  a  pair  of  straight  lines  at  right 
angles  to  each  other  ;  let  Fi  be  any  point  in  their  plane,  and 
draw  PiiHf  perpendicular  to  XX'. 

The  distances  OM  and  PiM  are  called  the  rectangular  co- 
ordinates of  Pi  with  reference  to  the  lines  XX'  and  YY'. 

OJf  is  called  the  abscissa,  and  FiMthe  ordinate;  the  lines 
XX'  and  YY'  are  called  the  axes  of  X  and  Y,  respectively, 
and  their  intersection  0  is  called  the  origin. 

29.  If  in  the  above  figure,  OM—  0N=  &,  and  the  per- 
pendiculars PiM=  P2N—  PzN=  P^M=  a,  the  points  Pi,  Pg? 
P3,  and  P4  will  have  the  same  co-ordinates. 

To  avoid  this  ambiguity,  the  following  conventions  have 
been  adopted : 

Abscissas  measured  to  the  right  of  0  are  considered  posi- 
tive., and  to  the  left.,  negative. 

Ordinates  measured  above  the  line  XX'  are  considered 
positive,^  and  below,  negative. 

Then  the  co-ordinates  of  the  four  points  will  be  : 


Point. 

Abscissa. 

Ordinate 

A 

b 

a 

P2 

-h 

a 

A 

-h 

—  a 

A 

b 

—  a 

APPLICATION   OF   ALGEBRAIC   SIGNS.  15 

Note.  In  the  figures  of  this  chapter,  the  small  letters  denote  the 
lengths  of  the  lines,  without  regard  to  their  algebraic  sign. 

30.  It  is  customary  to  denote  the  abscissa  and  ordinate  of 
a  point  by  the  letters  x  and  ?/,  respectively.  Thus  the  fact 
that  the  abscissa  of  a  point  is  equal  to  h  and  its  ordinate  to 
a,  is  expressed  by  the  saying  that  for  the  point  in  question 
x=h  and  y^=a. 

The  same  fact  ma}-  be  stated  more  concisely  by  referring 
to  the  point  as  "  the  point  (6,  a)",  where  the  first  quantity 
in  the  parenthesis  is  understood  to  be  the  abscissa,  and  the 
second  the  ordinate. 

31.  If  a  point  lies  upon  the  axis  of  X,  its  ordinate  is 
zero ;  and  the  same  is  true  of  the  abscissa  of  a  point  upon 
the  axis  of  Y. 

GENERAL    DEFINITIONS    OF    THE    FUNCTIONS. 

32.  We  will  now  give  general  delinitions  for  the  trigono- 
metric functions,  applicable  to  any  angle  whatever. 

Take  the  initial  line  as  the  positive  direction  of  the  axis  of 
X,  the  vertex  being  the  origin. 

Take  any  point  in  the  terminal  line,  and  construct  its  rec- 
tangular co-ordinates  by  dropping  a  perpendicular  to  the 
initial  line,  produced  if  necessary. 

Then,  designating  the  distance  of  the  assumed  point  from 
the  origin  as  the  "•  distance  "  of  the  point. 

The  SINE  is  the  ratio  of  the  ordinate  to  the  distance. 

The  cosine         is  the  ratio  of  the  abscissa  to  the  distance. 
The  tangent      is  the  ratio  of  the  ordinate  to  the  abscissa. 
The  COTANGENT  is  the  ratio  of  the  abscissa   to  the  ordinate. 
Tlie  secant         is  the  ratio  of  the  distance  to  the  abscissa. 
The  cosecant     is  the  ratio  of  the  distance  to  the  ordinate. 

Note.  These  definitions  include  those  of  Art.  10.  The  definitions 
of  the  versed  sine  and  coversed  sine,  given  in  Art.  11,  are  sufficiently 
general  to  apply  to  any  angle  whatever. 


16 


PLANE   TRIGONOMETRY. 


33.  We  will  now  apply  the  definitions  of  Ai't.  32  to  the 
angles  XOPj,  XOP^,  XOP^,  and  XOP^  in  the  following- 
figures  : 

Let  Pi,  Po,  P3,  and  P4  be  any  points  on  the  terminal  lines 
OPi,  OP21  OP^^  and  OP4,  and  let  their  co-ordinates  be  (6,  a), 
(—6,  ci),  (— ?>,  —  «),  and  (6,  —a),  respectively. 

Let  OPi  =  OPo  =  OPs  =  OP4  =  c. 

Then  by  the  definitions, 


sin  XOPi  =  -• 

c 

tanXOPi  =  -. 


cosXOPi  = 


cot  XOPi  =  -• 


sec  XOP.  =  -. 
^      5 


esc  XOPi  =  -. 

a 


sin  XOP2            =     -. 

c 

cosXOP2  = 

-b         b 
c            c 

tan  XOPo—    ^    —      ^, 

cot  XOP2  = 

-b         b 

—  5           5 

a           a 

-b          b 

CSCXOP2 

__     c 
a 

ArPLlCATIOX   OF   ALGEBRAIC    SIGNS. 


17 


X- 


PA-lj,<i) 


0 


sin  XOP,  = 
tan  XOI\  = 
8ecX0P3  = 


—  a_ 

c 

a 

c 

cos  XOP,=  ~^=      -. 

c             c 

—  u 

a 

Ij 

—  a          a 

-h 

c 

c 
b 

csoXOP,—    ^    —      ^ 

-b 

—  a         a 

P,(b,-a) 


sin  AOP4  = = 

c            c 

cos  XOP4            =     -• 

c 

6            b 

cotXOP^-— =  -- 
—  a          a 

sec  XOP4             =     -• 

b 

pqcXOP—    ^     —  -  /^ 

—  a          a 

34.  Since  the  terminal  lines  of  two  angles  which  differ  by 
a  mnltiple  of  360°  are  coincident  (Art.  27),  it  is  evident  that 
the  trigonometric  functions  of  two  such  angles  are  identical. 

Thus,  the  functions  of  50°,  410°,  770°,  —310°,  etc.,  are 
identical. 

It  is  customary  to  express  this  fact  by  saying  that  the 
trigonometric  functions  are  periodic. 


18 


PLANE   TRIGONOMETRY. 


35.  If  the  terminal  line  of  an  angle  lies  between  OX  and 
0  Y,  the  angle  is  said  to  be  in  the  first  quadrant ;  if  between 
Ol^and  OX',  in  the  second  quadrant;  between  OX'  and  0T\ 
in  the  third  quadrant;  between  OY^  and  OX,  in  the  fourth 
quadrant. 

Thus,  any  positive  angle  between  0°  and  90°,  or  between 
360°  and  450°,  or  any  negative  angle  between  —  270°  and 
—  360°,  is  in  the  first  quadrant. 

Any  positive  angle  between  90°  and  180°,  or  between  450° 
and  540°,  or  any  negative  angle  between  —  180°  and  —  270°, 
is  in  the. second  quadrant. 

36.  We  observe,  by  inspection  of  the  results  in  Art.  33, 
the  following  points  in  regard  to  the  algebraic  signs  of  the 
trigonometric  functions  in  the  different  quadrants  : 

For  any  angle  in  the  first  quadrant _  all  the  functions  are 
positive. 

In  the  second  quadrant^  the  sine  and  cosecant  are  positive., 
and  the  cosine.,  tangent.,  cotangent.,  and  secant  are  negative. 

In  the  third  quadrant,  the  tangent  and  cotangent  are  posi- 
tive, and  the  sine,  cosine,  secant,  and  cosecant  are  negative. 

In  the  fourth  quadrant,  the  cosine  and  secant  are  positive, 
and  the  sine,  tangent,  cotangent,  and  cosecant  are  negative. 

Zl.  It  is  customary  to  express  the  foregoing  principles  in 
tabular  form,  as  follows  : 


Functions. 

First 
Quad. 

Second 
Quad. 

Third 
Quad. 

Fourth 
Quad. 

Sine  and  cosecant   . 
Cosine  and  secant   . 
Tangent  and  cotangent     . 

+ 
+ 
+ 

+ 

+ 

+ 

APPLICATION  OF   ALGEBRAIC   SIGNS. 


19 


FUNCTIONS   OF  0°,  90°,  180°,  270°,  AND  360°. 
38.    To  find  the  functions  of  0°  and  360°. 


a   P(a,0) 


-X 


0 


The  terminal  line  of  0°  coincides  with  the  initial  line  OX. 
Let  P  be  a  point  on  OX,  such  that  OP  =  a . 
Then  by  Art.  31,  the  co-ordinates  of  P  are  (a,  0). 
Whence  by  definition. 


sinO°  =  -  =  0. 
a 

cosO°  =  -  =  l. 
a 


tanO°  =  ^=0. 
a 

cotO°  =  -  =  x 

0 


sec  0°  =  -^  =  1 

a 


CSC  0°  = 


00, 


By  Art.  34,  the  functions  of  360°  are  the  same  as  those  of  0°. 
39.    To  find  the  functions  of  90°» 


] 

r 

P  (0,a) 

a 

^90* 

V" 

0 

JL 

Let  P  be  a  point  on  0  Y,  such  that  OP  =  a. 
Then  the  co-ordinates  of  P  are  (0,  a). 
Whence  by  definition. 


sin  90°  =  - =  1. 

a 


cos  90°  =  -  =  0. 

a 


tan  90°  =  -=00. 
0 


cot  90°  =  -=0. 
a 


a 


sec  90°  =  -  =  00. 

0 

csc90°  =  - =  1. 
a 


20 


PLANE    TRIGONOMETRY. 


40.    To  find  the  functions  of  180°. 


-X 


P  (-rt,  0) 


o 


Let  P  be  a  point  on  0X\  such  that  0P=  a. 
Then  the  co-ordinates  of  P  are  (  —  a,  0). 
Whence  by  definition, 


sin  180°  =  -      =0. 
a 


tan  180°  = =  0. 

—  a 

sec  180°  =  -^  = -1- 

—  a 


fi 

cos  180°  = =-1. 

a 

cot  180°  =:=l^=oo. 


CSC  180°  =  -      =00. 
0 


41.    To  find  the  functions  of  270°. 


270 


Y' 


Let  P  be  a  point  on  OF',  such  that  0P=  a. 
Then  the  co-ordinates  of  P  are  (0,  —  a). 
Whence  by  definition, 


__  /-» 

sin  270°  = =-1. 

a 

tan  270°  =  ^^=00. 


cos  270° 


0 


a 


=  0. 


cot  270°  = =0. 

—  a 


sec  2' 


a 


—  00. 


CSC  270°  = 


a 


-1, 


c» 


APPLICATIOX   OF   ALGEBRAIC    SIGXS. 


21 


Note.  No  absolute  meaning  can  be  attached  to  such  results  as 
cot  0'-'  =  CO,  tan  90°  =  x ,  etc.  The  equation  cot  0°  =  ao  merely  signifies 
that  as  an  angle  approaches  0°  as  a  limit,  its  cotangent  increases  with- 
out limit. 

A  similar  interpretation  nmst  be  given  to  the  equations  esc  0°  =  :o, 
sec  90°  —  X,  etc. 

FUNCTIONS   OF   (-.1)   IN   TERMS   OF   THOSE   OF   A. 
42.    To  prove  the  formulae^ 

sin(— ^1)  =  —  sin^,        cos(  — .1)=      cos^l, 
tan(  — yl)=  — tan^, 
sec(— ^1)=      sec^, 
for  any  value  of  A. 


cot  {  —  A)=  —  cot  ^1 , 

CSC(— ^1)=  —CSC  .4, 


(9) 


There  will  be  four  cases,  according  as  A  is  in  the  first, 
second,  third,  or  fourth  quadrant. 

In  each  figure,  let  the  positive  angle  XOP  (indicated  by 
the  fill  arc)  represent  the  angle  A^  and  the  negative  angle 
XOP'  (indicated  by  the  dotted  arc)  the  angle  (  —  A). 

Draw  J^F'  perpendicular  to  XX' ;  then  the  right  triangles 
07*3/ and  OP'Jf  have  the  side  03/ and  the  angle  POM  of 
one  equal  to  the  side  03/ and  the  angle  P'OMoi  the  other, 
and  are  equal. 

Whence,  PM=  P'3/and  0P=  OP'. 

Let  P3I=  P'M=  a,  0M=  b,  and  0P=  OP'  =  c. 

Cask  I.    A  in  the  first  quadrant^  —A  in  the  fourth. 


P(b,d) 


22  PLANE   TRIGONOMETRY. 

By  definition,  we  have  : 


Angle. 

Sin. 

Cos. 

Tan. 

Cot. 

Sec. 

Csc. 

A 

a 

b 

a 

b 

C 

C 

c 

c 

b 

a 

b 

a 

-A 

a 

b 

a 

b 

c 

c 

c 

c 

b 

a 

b 

a 

It  is  evident  from  the  above  that  the  cosine  and  secant  of 
—  A  are  equal  respectively  to  the  cosine  and  secant  of  A, 
while  the  sine,  tangent,  cotangent,  and  cosecant  of  —A  are 
equal  respectively  to  the  negatives  of  the  sine,  tangent,  co- 
tangent, and  cosecant  of  A. 

Whence  we  obtain  the  formulae  (9). 

Case  II.    A  in  the  second  quadrant^  —A  in  the  third, 

Y 

P  (-h,a) 


PX-h,-a) 


Angle. 

Sin. 

Cos. 

Tan. 

Cot. 

Sec. 

Csc. 

A 

a 
c 

b 

c 

a 
~b 

b 
a 

C 

b 

c 
a 

-A 

a 

c 

b 

c 

a 
b 

b 

a 

c 
~b 

c 

a 

Whence  we  obtain  the  formulae  (9)  as  before. 


APPLIOATIOX   OF    ALGEBRAIC   SIGNS. 


^S 


Case  III.    A  in  the  third  quadrant,  —A  in  the  second. 

Y 


P  {-h,-a) 


Angle. 

Sin. 

Cos. 

Ta  n . 

Cut. 

Stc. 

Csc. 

A 

a 

b 

a 

b 

c 

C 

c 

C 

b 

a 

b 

a 

a 

h 

a 

b 

c 

■     c 

—A 

c 

C 

b 

a 

b 

a 

Whence  we  ol)tiiin  the  formiih^  (9)  as  before. 

Case  IV.    A  in  the  fourth  quadrant,  —A  in  the  first. 

Y 


Angle. 

Si7l. 

Cos. 

Tan. 

Cot. 

Sec. 

Csc. 

A 

a 

b 

a 

b 

C 

C 

c 

c 

b 

a 

b 

a 

-A 

a 

b 

a 

b 

c 

c 



— 

— 

— 

— 

c 

c 

b 

ft 

b 

a 

Whence  we  obtain  the  formulae  (9)  as  before. 


24 


PLANE   TRIGOK^OMETRY. 


43.  The  results  of  Art.  42  may  be  stated  as  follows : 

The  sine,  cosine,  tangent,  cotangent,  secant,  and  cosecant  of 
a  negative  angle  are  equal  respectively  to  the  negative  sine,  the 
cosine,  the  negative  ta7igent,  the  negative  cotangent,  the  secant, 
and  the  negative  cosecant,  of  the  absolute  value  of  the  angle. 

FUNCTIONS   OF  (90°  +  ^)  IN  TERMS   OF   THOSE   OF  A. 

44.  To  prove  the  forniulm, 

sin(90°  +  ^)  ==      cos^,  cos(90°  +  A)=-  s\\yA,  ' 

cot(90°  +  ^4)=  -tan^, 
csc(90°  +  ^)=      sec^,  ^ 


(10) 


tan(90"  +  ^)=  -cot.4, 

sec(90°  +  ^)=  -csc^, 
for  any  value  of  A. 

There  will  be  four  cases,  according  as  A  is  in  the  first, 
second,  third,  or  fourth  quadrant. 

In  each  figure,  let  the  positive  angle  XOP  (indicated  by 
the  full  arc)  represent  the  angle  A,  and  the  positive  angle 
XOP'  (indicated  by  the  dotted  arc)  the  angle  (90°  +  ^). 

Lay  off  0P=  OP',  and  draw  PJf  and  P'M'  perpendicular 
to  XX'. 

Since  OP  and  03f  are  perpendicular  to  OP'  and  P'M' 
respectively,  the  angles  POilf  and  OP'M'  are  equal. 

Then  the  right  triangles  0PM  and  OP'M'  have  the  hypote- 
nuse OP  and  the  angle  POM  of  one  equal  to  the  hypotenuse 
OP'  and  the  angle  OP'M'  of  the  other,  and  are  equal. 

Whence,  P3I=  OM'  and  0M=  P'M'. 

Let  PM=  OM'  =  a,  0M=  P'M'  =  b,  and  0P=  OP'  =  c. 

Case  I.    A  in  the  first  quadrant,  90°  +  A  m  the  second. 


APPLICATIOX   OF   ALGEBRAIC   SIGNS.  25 

By  definition,  we  have  : 


Angle. 

Sin. 

Cos. 

Tan. 

Cot. 

Sec. 

Csc. 

A 

a 

h 

a 

b 

C 

c 

c 

c 

b 

a 

b 

a 

90°+^ 

h 

a 

b 

a 

c 

c 

c 

c 

a 

b 

a 

b 

It  is  evident  from  the  above  that  the  sine  and  cosecant  of 
90°  +  A  are  equal  respectively  to  the  cosine  and  secant  of  A, 
while  tire  cosine,  tanoont,  cotangent,  and  secant  of  90°+^ 
are  equal  respectively  to  the  negatives  of  the  sine,  cotangent, 
tangent,  and  cosecant  of  A. 

Whence  we  obtain  the  formuhe  (lo) . 

Case  II.   A  in  the  second  quadrant,  90°  ^  A  in  the  third. 


^X 


P  (-a,-h) 


Angle. 

Sin . 

Cos. 

Tan. 

Cot. 

Sec. 

Csc. 

A 

a 

b 

a 

b 

C 

c 

— 

— 

— 

— 

— 



c 

c 

b 

a 

b 

a 

90°+^ 

b 

a 

b 



a 

c 

c 

c 

c 

a 

b 

a 

b 

Whence  we  obtain  the  formuljs  (lo)  as  before. 


PLANE   TRIGONOMETRY. 
Case  III.    A  in  the  third  quadrant^  90°  +  ^  in  the  fourth. 


P  i-h,-a) 


P  (a,-h) 


Angle. 

Sin. 

Cos. 

Tail. 

Cot. 

Sec. 

Csc. 

A 

a 

b 

a 

b 

C 

c 

c 

c 

b 

a 

b 

a 

90°+^ 

b 

a 

b 

a 

c 

c 

c 

c 

a 

b 

a 

b 

Whence  we  obtain  the  formulae  (lO)  as  before. 

Case  IV.    A  in  the  fourth  quadrant,  90°  +  ^  in  the  first. 

p\a,h) 


P  (b,-a) 


Angle. 

Sin. 

Cos. 

Tan. 

Cot. 

Sec. 

Csc. 

A 

a 

b 

a 

b 

C 

c 

c 

c 

b 

a 

b 

a 

90°+^ 

b 

a 

b 

a 

c 

c 

c 

c 

a 

b 

a 

b 

Whence  we  obtain  the  formulae  (lO)  as  before. 


APPLICATION  OF  ALGEBRAIC   SIGXS.  27 

45.  The  results  of  Art.  44  may  be  stated  in  the  follow- 
ing form  : 

The  sine,  cosine,  tangent,  cotangent,  secant,  and  cosecant  of 
any  angle  are  equal  respectively  to  the  cosine,  the  negative  sine, 
the  negative  cotangent,  the  negative  tangent,  the  negative  cose- 
cant, and  the  secant,  of  an  angle  90°  less. 

46.  To  find  the  values  of  the  functions  of  90°  —  A  in 
terms  of  those  of  A. 

By  Arts.  45  and  42,  we  have : 

sin  (90°-  .1)  =  cos  ( -.^1)  =  cos  A. 
cos(00°-yl)=  _sin(-.l)=  sin^. 
tan  (90°-  A)  =  -cot  ( -^1)  =  cot  A. 
cot  (90°-. 4)=  -tan(-^l)  =  tan^. 
sec(90°-^)=  -csc(-^)=  csc^. 
csc(90°-.4)=      sec(-.d)=  sec^. 

We  liave  thus  proved  the  formulae  of  Art.  14  for  any 
value  of  A. 

47.  To  find  the  vcdues  of  the  functions  of  180°— yl  in 
terms  of  those  of  A. 

By  Arts.  45  and  4G  we  have  : 

sin  (180°-^)=  cos  (90°-^)=  sin^. 
cos  (180°-  .4)  =  -  sin  (90°-  A)  =  -  cos  A. 
tan  (180°-  A)  =  -  cot  (90°-  A)  =  -tan  A. 
cot  (180°-  A)  =  -  tan(90°-  A)  =  -  cot  A. 
sec  (180°-  ^)  =  -  csc(90°-  A)  =  -  sec  A. 
CSC  (180°-. 4)=      sec(90°-^)=      csc^. 

Since  the  angles  A  and  180°— J.  are  supplements  of  each 
other,  these  formulae  express  the  values  of  the  functions  of 
the  supplement  of  an  angle  in  terms  of  those  of  the  angle 
itself. 


28  PLANE   TRIGONOMETRY. 

48.  To  find  the  values  of  the  functions  of  180°  +  A  in 
terms  of  those  of  A. 

By  Arts.  45  and  44,  we  have  : 

sin  (180°+^)=  cos  (90°+^)= -sin ^.  . 
cos  (180°+^)=  -sin  (90°+^)=  -cos  J.. 
tan  (180°+  A)=-  cot  (90°+  ^)  =  tan  A. 
cot  (180°+^)=  -tan  (90°+^)=  cot^. 
sec  (180°+^)=  -esc  (90°+^)=  -sec^. 
CSC  (180°+^)=      sec  (90°+^!)= -CSC  A 

49.  To  find  the  values  of  the  f  mictions  of  270°— -4  in  terms 
of  those  of  A. 

By  Arts.  45  and  47,  we  have  : 

sin  (270°-^)=  cos  (180°-^)  =  - cos ^. 
cos  (270°-^)  =  -  sin  (180°- J.)  =  -  sin^. 
tan  (270°-  ^)  =  -  cot  (180°-  ^.)  =  Got  A. 
cot  (270°-  A)  =  -  tan  (180°-  A)  =  tan  A. 
sec  (270°-^)  =  -  CSC  (180°- ^)  =  -  csc^. 
CSC  (270°-  ^)  =      sec  (180°-  A)  =  -  sec  A, 

50.  To  find  the  values  of  the  functions  of  270°+  ^  in  terms 
of  those  of  A. 

By  Arts.  45  and  48,  we  have : 

sin  (270°+  A)  =      cos  (180°+  A)  =  -  cos  A. 

cos  (270°  +  ^)  =  -  sin  (180°+^)=      sin  ^. 

tan  (270°  +  ^)  =  -  cot  (180°+  J.)  =  -  cot  A. 

cot  (270°+^)^ -tan  (180°+ ^)  =  -  tan.4. 

sec  (270°+^!)  = -CSC  (180°  +  ^)=      csc^. 

CSC  (270°+.4)=      sec  (180°+. 4)  =  - sec ^. 


applicatio:n"  of  algebraic  SIGXS.  29 

51.  To  Jind  the  values  of  the  functions  of  360°—^  and 
360°+^  in  terms  of  those  of  A. 

By  Art.  34,  the  functions  of  3G0°— ^4  are  the  same  as 
those  of  —A^  and  the  functions  of  360°+  --i  are  the  same  as 
those  of  A. 

Whence  by  Art.  42, 
sin  (S(')0^-A)  =  -  sin  A.  cos  (3G0°-^)=      cos^. 

tan  (360°-^)  =  -tau^.  cot  (360°-^)  =  -  cot.4. 

sec  (360°-^!)=     sec  A  esc  (300°-^)  =  - esc  ^. 

And, 

sin  (360° 4- ^4)=      sin.4.  cos  (360°  + ^1)  =      cos^. 

tan  (360°+  A)  =      tun  A.  cot  (360°+  ^1)  =      cot^. 

sec  (360°+^)=      sec^.  esc  (360°  +  ^)=      csc^. 

52.  Tiie  operation  illustrated  in  Art.  51  may  be  extended 
indefinitely  ;  we  should  find  for  the  angles  450°-  A,  810°-^, 
—  270°—  A,  etc.,  the  same  results  as  for  1)0°— ^rl ;  and  so  on. 

53.  The  results  of  Arts.  42  to  52  may  be  expressed  in  the 
following  rule,  which  is  derived  by  inspection  from  the  for- 
muh^  of  Arts.  42  to  51  : 

Any  function  of  0°,  or  an  even  multiple  of  90°,  i^lus  or 
minus  A.  is  the  sx'me  function  of  A  ;  and  any  function  of  an 
ODD  midtiple  of  90°,  plus  or  minus  A^  is  the  complementary 
function  of  A. 

To  obtain  the  algebraic  sign,  regard  A  as  an  acute  angle, 
and  apply  the  table  of  Art.  37. 

1.    Required  the  value  of  sec  (990°+^). 

Since  990°  is  an  odd  multiple  of  90°,  the  absolute  value  of 
the  result  is  esc  ^4. 

And  if  A  is  an  acute  angle,  990°  +  ^  is  in  the  fourth  quad- 
rant, in  which  the  sign  of  the  secant  is  positive. 

Hence,  sec  (990°+  .4)  =  esc  JL. 


30  PLANE   TRIGONOMETRY. 

2.  Required  the  value  of  tan  (-  180°-  A) . 

Since  180°  is  an  even  multiple  of  90°,  the  absolute  value  of 
the  result  is  tan^. 

And  if  A  is  an  acute  angle,  —  180°—^  is  in  the  second 
quadrant,  in  which  the  sign  of  the  tangent  is  negative. 

Hence,  tan  (-  180°-  ^)  =  -  tan^. 

EXAMPLES. 

Find  the  values  of  the  functions  of  the  following  angles  in 
terms  of  the  functions  of  A : 

3.  450°- A  7.      630°- A         11.   -180°+ A 

4.  450°  +  ^.  8.      900°-^.         12.   -180°- A 

5.  540°-^.  9.   -90°+^.         13.   -270° 4-^. 

6.  540°+^.         10.   -90°-^.         14.   -720°  +  ^. 

54.  The  functions  of  any  positive  or  negative  angle  what- 
ever may  be  expressed  in  terms  of  the  functions  of  an  angle 
between  0°  and  90°. 

1.  Express  sin  317°  as  a  function  of  an  angle  between  0° 
and  90°. 

We  have  by  the  rule  of  Art.  53, 

sin  317°=  sin  (270°+  47°)  =  -  cos  47°, 
or  sin  317°=  sin  (360^-43°)  =  -  sin  43°. 

EXAMPLES. 

Express  the  following  in  terms  of  the  functions  of  angles 
between  0°  and  90° : 

2.  cos  152°.  4.    sec (-77°).         6.    cot  (-129°). 

3.  tan-522°.  5.    esc  230°.  7.    sin  865°. 

Express  the  following  in  terms  of  the  functions  of  angles 
between  0°  and  45° : 

8.  cot    83°.  10.    sec  165°.  12.    tan  520°. 

9.  sin  (-50°).       11.    cos  (-303°).  13.    esc  768°. 


APPLICATION   OF   ALGEBRAIC   SIGNS. 


31 


55.  The  following  table  gives,  for  convenient  reference, 
the  functions  of  the  angles  0°,  30°,  45°,  60°,  90°,  120°,  etc. 

The  values  of  the  functions  of  0°,  30°,  45°,  60°,  90°,  180°, 
270°,  and  360°  have  already  been  proved  in  Arts.  16,  17,  38., 
39,  40,  and  41  ;  the  others  may  be  derived  b}"  aid  of  Art. 
53,  and  are  left  as  exercises  for  the  student. 

As  an  illustration,  we  will  give  the  proof  for  cot  150°. 

By  the  rule  of  Art.  53  : 

cotl50°=cot(180°-30°)=- cot30°=  -V^   (Art.  17). 


Angle. 

Sin. 

Cos. 

Tnn. 

Cot. 

Sec. 

CV. 

0° 

0 

1 

0 

00 

1 

00 

30° 

i 

iy/s 

^v^ 

V5 

Iv^ 

2 

45° 

iv/2 

iv/2 

1 

1 

Vi 

V/2 

60° 

i^S 

i 

v^ 

iv'3 

2 

lv/3 

90° 

1 

0 

00 

0 

00 

1 

120° 

iv^ 

-i 

-V3 

-iv/3 

-2 

lv/3 

135° 

iV2 

-i\/~2 

-1 

-1 

-\/2 

V/2 

150° 

i 

-ifd 

-\fi 

-V/3 

-fV3 

2 

180° 

0 

-1 

0 

00 

-1 

00 

210° 

-i 

-iy/s 

-W3 

S/3 

-In/3 

-2 

225° 

-iv/2 

-^^2 

1 

1 

-V/2 

-V2 

240° 

-i\/3 

-i 

V/3 

iv^ 

-2 

-lv/3 

270° 

-1 

0 

00 

0 

00 

-1 

300° 

-iy^ 

i 

-v/5 

-iV3 

2 

-lv/3 

315° 

-i\/2 

isTi 

-1 

-1 

V2 

-v^ 

330° 

-i 

i\/n 

-iV3 

-sft 

IV3 

-2 

360° 

0 

1 

0 

00 

1 

00 

32 


PLANE   TEIGOXOMETPvY. 


56.     Given  tlie  value  of  one  of  the  functions  of  an  angle,  to 
find  the  values  of  the  remaining  functions.    (Compare  Art.  15.) 

3 
1.    Given  sin  ^  =  —  - ;  requii'ed  the  values  of  the  remain- 

iug  functions  of  A. 

The  example  may  be  solved  by  a  method  similar  to  that 
employed  in  Art.  15. 

Since  the  sine  is  the  ratio  of  the  ordinate  to  the  distance, 
we  may  regard  the  poiut  of  reference  as  having  its  ordinate 
equal  to  —3,  and  its  distance  equal  to  5. 

There  are  two  such  points,  P  and  P',  and  consequently  two 
angles,  XOP  and  XOP',  in  the  third  and  fourth  quadrants 
respectively,  either  of  which  satisfies  the  given  condition. 


Then   031=  OM'  =  \^ OP''  -  PM''  =  V^25  -  9  =  4,  and  the 
co-ordinates  of  P  are  (  —  4,  —  3),  and  of  P',  (4,  —  3). 


Whence,  cos  XOP= 5 

'  5 


cos  XOP'  = 


tan  XOP  =      -^ 
4 


tan  XOP'  =  -  -, 
,4 


cot  XOP  =       -\ 
3 


cot  XOP' =  --, 
3 


sec  XOP  =  --, 
4 


sec  XOP'  =      -, 
4 


esc  XOP  =  --, 
3 


CSC  XOP'  =  -  -. 
3 


APPLICATION   OF   ALGEBRAIC    SIGXS. 


33 


Thus  the  two  solutions  to  the  exaraple  are 


cos  ^  =  q:  -•      tan  ^  =  ± - ■ 


cot  ^1  =  ±  -? 
3 


sec^l=  q:-?      CSC  ^4= - 

4  3 

where  the  upper  signs  refer  to  the  angle  XOP^  and  the  lower 

signs  to  XOP'. 

2.    Criven  cot  ^1  =  -^  ;  requii'ed  the  values  of  the  remain- 
ing functions  of  A. 

The  equation  may  be  written  cot  A  = i^- 

We  may  then  regard  the  point  of  reference  as  having  its 

abscissa  equal  to  12  and  ils  ordinate  equal  to  5,  or  as  having 

its  abscissa  equal  to  —12  and  its  ordinate  equal  to  —  5  ;  and 

there  are  consequently  ^?ro  aiu/Jes,  XOP  und  XOP',  in  the  first 

and  third  (piadrants  respectively,  either  of  which  satisfies  the 

given  condition. 

Y 


Then  0P=  OP'  =  sl OM' -[- P][f  =  v/l44  +  25  =  13. 


Whence,  sin  XOP=  -^5 

13 


cos  XOP  = 


12 
13' 


tanXOP  =  — . 
12 


sec  XOP  = 
CSC  XOP  = 


13 

12' 

13 


sin  XOP'  =  -  — , 
13 

cos  XOP'  =  -  — , 
13 

tan  XOP'-      — , 
12 

sec  XOP' =  -— » 
12 

CSC  XOP  =  -— . 


34  PLANE   TRIGONOMETRY. 


Thus  the  two  solutions  are 


5  12  5 

sin^  =  ± — 7     cos  ^  =  ± — 1     tan^  =  — ? 
13  13  12 

sec^  =  ± — •»      csc^  =  ±  — 
12  5 


EXAMPLES. 
In  each  case  find  the  values  of  the  remaining  functions : 

3.  sin^  =  i.  7.    sec^  =  -.  11.    cos^  =  --. 

4  3  6 

4.  cot ^  =  2.  8.    cos^= 12.    sin^  =  .T. 

2 

5.  csc^=---  9.   cscJ.  =  V^.        13.   cot^==i. 

2  X 


6.    tan^=-— •       10.    tan^=2V^2.      14.    sec^  = 


15  a 

PROOFS   OF  THE  FUNDAMENTAL    FORMULA    FOR    ANY 

ANGLE. 

57.    We  have  from  the  general  definitions  of  Art.  32, 

.  ,      ordinate  ,  distance      ., 

sm  A  X  CSC  A  = X  — =  1 . 

distance     ordmate 

4  A      abscissa     distance      ^ 

cos  Ax  sec  ^  =  X =  1 . 

distance     abscissa 

,       A  A.  A      ordinate     abscissa      ^ 

tan  J.  X  cot  ^  = X =  1 . 

abscissa     ordinate 

Whence  we  obtain,  as  in  Art.  18, 

111 

sin^  = •)        tanJ[= -5        sec^= -•> 

CSC  A  cot  A  cos  A 

cos  A  = ^        cot  A  = 1        esc  A  = -• 

sec  A  tan  A  sm  A 


APPLICATIOX   OF   ALGEBRAIC    SIGNS.  35 

58.    To  prove  the  formulae,, 

bin-^1  +  cos^.l  =  1 ,    sec- .4  =  1  +  tan-^,  csc^^  =  1  +  cot- .4, 

for  any  value  of  A. 

Since  the  distance  of  the  point  of  reference  is  the  hypote- 
nuse of  a  riglit  triangle  wliose  sides  are  equal  to  the  absolute 
values  of  the  abscissa  and  ordinate,  we  have  by  Geometry, 

(ordinate)- +  (abscissa)- =  (distance)-. 

This  ma}'  be  written  in  the  forms, 

/ordinateV      /absciss aY_  ^ 
\distancey       \distance/ 

/distance Y_  -,       /ordinaceV 
\abscissay  \abscissa/ 

and  /'distanceY-_  ,       /abscissa^^ 


ordinate/  \ordinatey 

Hence  for  any  value  of  A,  we  have 
sin- .4  +  cos-yl  =  1 ,   sec- .4  =  1  +  tan-yl,    csc-yl  =  1  +  cotM. 

59.    To  prove  the  formulae 

4      sin  A        J      .J      cos  A 
tan  A  = ana  cot  A  = -i 

cos  A  sin  A 

for  any  value  of  A. 

By  definition,  ordinate 

.      ordinate      distance      sin  A 

tan  A  = =  -; — : = -> 

abscissa      abscissa      cos  J. 

distance 

and  abscissa 

,    .       abscissa      distance      cos^ 

cot  A  = =  — -. — —  = -• 

ordinate      ordinate      sin  A 

distance 


36 


PLANE   TRIGONOMETRY. 


IV.    MISCELLANEOUS  THEOREMS. 

TO  EXPRESS  EACH  OF  THE  SIX  PRINCIPAL  FUNCTIONS 
IN  TERMS   OF   THE   OTHER  FIVE. 

60.    The  following  table  expresses  the  value  of  each  of 
the  six  principal  functions  in  terms  of  the  other  five  : 


sin 

COS 

tan 

cot 
sec 

CSC 

tan 

1 

y/l  +  cot- 
cot 

1 

CSC 

y/sec2  - 
sec 

1 

sec 

1 

\/l  -  cos2 

y/l  —  cos- 
cos 

cos 

y/l  +  tan- 

1 

y/csc-  —  1 

CSC 

1 

y/l  -  sin^ 

sin 
\/l  —  sin--2 

sjl  +  tan^ 

1 
tan 

y/l  +  cot2 

1 
cot 

y/l  +  cot2 
cot 

y/sec"-2  - 
1 

1 

y/csc'-  —  1 

y/l  —  sin- 
sin 

1 

y/csc-  —  1 

CSC 

y/l  -  cos2 

1 
COS 

1 

y/l  —  cos- 

y/sec2  - 
sec 

a 

y/l  +  tan"-^ 

y/l  —  sin^ 

1 

sin 

y/csc'-  —  1 

y^l  +  tan- 
tan 

y/l  +  cot- 

y/sec^  - 

^ 

The  reciprocal  forms  were  proved  in  Art.  57  ;  the  others 
may  be  derived  by  aid  of  Arts.  57,  58,  and  59,  and  are  left 
as  exercises  for  the  student. 

As  an  illustration,  we  will  give  a  proof  of  the  formula 

.      Vcsc^  A-1 

cos  A  = ' 

CSC  A 

By  Art.  58,  cos  A=\/l-  s'm^A 


=  i  1 


1      ^  y/csc^  A-l 
csc^  A  CSC  A 


MISCELLAXEOUS   THEOREMS. 


3T 


LIMITING  VALUES    OF    ^^^   AND    ^^5^- 


61.    To  find  the  limiting  values  of  the  fractions 


tan  X 


X 


sin  X 

X 


arid 


as  X  approaches  the  limit  0. 


Note.     AVe  suppose  x  to  be  expressed  in  circular  measure  (Art.  4). 


Let  OPXP'  be  a  circular  sector;  draw  PT  and  P'T  tan- 
gent to  the  arc  at  P  and  P',  and  join  OT  and  PP'. 
By  Geometry,  PT=P'T. 

Then  OT  is  perpendicuhir  to  PP'  at  its  middle  point  M, 
and  bisects  the  arc  PP'  at  X. 

Let  Z  XOP=Z  XOP'  =  X. 

By  Geometry,      arc  PP'  >  chord  PP',  and  <  PTP. 

Whence,  arc  PX  >  PJ/,    and  <  PT. 


Therefore. 


arc  PX  ^  PM         .       PT 
OP  OP  OP 


or,  by  Art.  5,    ciic.  meas.  x  >  sin.i-,  and  <  tanoj.  (a) 

Representing  the  circular  measure  of  x  by  x  simply,  and 
dividing  through  by  sin  x,  we  have 

1 


•'^'      ^  1          1    ^  tan  X 
>  1,  and  < or 


That  is. 


sin  X 
sin  X 

X 


sm  X       cos  X 


<  1 ,  and  >  cos  x. 


38 


PLANE   TRIGONOMETRY. 


But  as  X  approaches  the  limit  0,  cos  x  approaches  the  hmit 
1  (Art.  38). 

Hence, approaches  the  limit  1  as  i»  approaches  0. 

X 

tana;        sin  a;        since  1 

X 


Affain, 

°  XX  cos  X         X         cos  X 

But  each  factor  approaches  the  limit  1  as  a;  approaches  0. 
Hence,  -^ —  approaches  the  limit  1  as  a;  approaches  0. 

X 


LINE  VALUES   OF  THE   TRIGONOMETRIC   FUNCTIONS. 
62.    Let  AOB  be  any  angle. 


With  0  as  a  centre,  and  a  radius  equal  to  1,  describe  the 
circle  AB;  draw  BD  and  AE  perpendicular  to  XX',  and 
CF  perpendicular  to  YY'. 

Then  by  Art.  32,  the  functions  of  AOB  are  : 


MISCELLANEOUS   THEOREMS. 


39 


;  Si'n.      1  Cos. 

1        1 

Tan. 

Cot. 

Sec.            Csc. 

Fig.  1 
Fig.  2 
Fig.  3 
Fig.  4 

BD 
OB 

BD 
OB 

BD 
OB 

BD 
OB 

OD 
OB 

OD 
OB 

OD 
OB 

OD 
OB 

BD 
OD 
BD 
OD 

BD 
OD 

BD 
OD 

1 

bobobobo 
bbbbbbbb 

OB    \      OB 
OD         BD 

OB    j  OB 
OD    '  BD 

OB          OB 
OD         BD 

OB          OB 
OD         BD 

P>ut  since  the  right  triangles  OBD^   OEA,  and  OCF  are 
similar,  and  OA  =  OC  =  1 ,  we  have, 


BD^AE 
OD  ~  OA 


AE, 


OB^OE 

OD  ~  OA 


OE, 


Qll=.9l=CF 
BD      OC 


BD      OC 


Whence,  since  0B=  1,  the  fnnctions  of  AOB  are 


Sin. 

Cos. 

Tan. 

Cot. 

Sec. 

Csc. 

Fig.  1 

BD 

OD 

AE 

CF 

OE 

OF 

Fig.  2 

BD 

-OD 

-AE 

-GF 

-OE 

OF 

Fig.  3 

-BD 

-OD 

AE 

CF 

-  OE 

-OF 

Fig.  4 

-BD 

OD 

-AE 

-CF 

OE 

-  OF 

That  is,  if  the  radius  of  the  circle  is  unity, 

The  sine  is  the  perpendicular  drawn  to  XX'  from  the 
intersection  of  the  circle  with  the  terminal  line. 

The  cosine  is  the  line  drawn  from  the  centre  to  the  foot 
of  the  sine. 

The  tangent  is  that  portion  of  the  geometrical  tangent  to 
the  circle  at  its  intersection  with  OX  included  between  OX 
and  the  terminal  line,  produced  if  necessary. 


40 


PLANE   THIGOXOMETRY. 


The  cotangent  is  that  portion  of  the  geometrical  tangent  to 
the  circle  at  its  intersection  with  0  Y  included  between  0  Y 
and  the  terminal  line,  produced  if  necessary. 

The  secant  is  that  portion  of  the  terminal  line,  or  terminal 
line  produced,  included  between  the  centre  and  tlie  tangent. 

The  cosecant  is  that  portion  of  the  terminal  line,  or  terminal 
line  produced,  included  between  the  centre  and  the  cotangent. 

And  with  regard  to  algebraic  signs, 

Sines  and  tangents  measured  above  XX'  are  positive^  and 
beloiv,  negative  ;  cosines  and  cotangents  measured  to  the 
right  of  YY'  are  positice^  and  to  the  left^  negative;  secants 
and  cosecants  measm'ed  on  the  terminal  line  itself  are  posi- 
tive^ and  on  the  terminal  line  produced,  negative. 

63.  The  above  are  called  the  line  vcdues  of  the  trigono- 
metric functions.  They  simply  represent  the  values  of  the 
functions  when  the  radius  is  unity ;  that  is,  the  numerical 
value  of  the  sine  of  an  angle  is  the  same  as  the  number 
which  expresses  the  length  of  the  perpendicular  drawn  to 
XX'  from  the  intersection  of  the  circle  and  terminal  line. 


64.    To  trace  the  changes  in  the  six  principcd  trigonometric 
functions  of  an  angle  as  the  angle  increases  from  0°  to  360°. 


Let  the  terminal  line  start  from  the  position  OA,  and 
revolve  about  0  as  a  pivot  in  the  direction  of  0 1",  occupying 
successively  the  positions  OB^.  OB...  OC,  OB^,  OB^,  etc. 


MISCELLANEOUS   THEOREMS.  41 

Then  since  the  sine  of  the  angle  commences  with  the  vakie 
0,  and  assumes  in  succession  the  values  ByD^^  BoDo,  OC, 
B^D^,  B^D^^  etc.  (Art.  GO),  it  is  evident  that  as  the  angle 
increases  from  O''  to  90°,  the  sine  increases  fron  0  to  1  ;  from 
90°  to  180°,  it  decreases  from  1  to  0  :  from  180°  to  270°,  it 
decreases  (algebraically)  from  0  to  —1;  and  from  270°  to 
360°,  it  increases  from  —1   to  0. 

Since  the  cosine  commences  with  the  value  OA^  and  as- 
sumes in  succession  the  values  OZ>i,  OD.j-,  0,  —01);^^,  —OD^, 
etc.,  from  0°  to  90°,  it  decreases  from  1  to  0  ;  from  90°  to 
180°,  it  decreases  from  0  to  -1  ;  from  180°  to  270°,  it  in- 
creases from  —1  to  0  ;  and  from  270°  to  360°,  it  increases 
from  0  to  1. 

Since  the  tangent  commences  with  the  value  0,  and  as- 
sumes  in   succession    the    values    ^-LE'i,    AE.2-,   ^c,    —  vLEg, 

—  AE^,  etc.,  from  0°  to  90°,  it  increases  from  0  to  x  ;  from 
90°  to  180°,  it  increases  from  —  x  to  0;  from  180°  to  270°, 
it  increases  from  0  to  oo  ;  and  from  270°  to  360°,  it  in- 
creases from  —  X  to  0. 

Since  the  cotangent  commences  at  x,  and  assumes  in  suc- 
cession the  values  CFy,  CR^,  0,  -OT;,  —CF^,  etc.,  from  0° 
to  90°,  it  decreases  from  x  to  0 ;  from  90°  to  180°,  it  de- 
creases from  0  to  —  x;  from  180°  to  270°,  it  decreases  from 
X  to  0  ;  and  from  270°  to  360°,  it  decreases  from  0  to  —  x. 

Since  the  secant  commences  at  OA,  and  assumes  in  suc- 
cession the  values  OE^^  OE.,,  x,  —OE.,  —OE^,  etc.,  from  0° 
to  90°,  it  increases  from  1  to  x  ;  from  90°  to  180°,  it  increases 
from  —X  to  —  1  ;  from  180°  to  270°,  it  decreases  from  —1 
to  —  X  ;  and  from  270°  to  360°,  it  decreases  from  x  to  1. 

Since  the  cosecant  commences  at  x,  and  assumes  in  suc- 
cession the  values  OF,,  OF.,  OC,  OF^,  OF^,  etc.,  from  0°  to 
90°,  it  decreases  from  x  to  1  ;  from  90°  to  180°,  it  increases 
from  1  to  X  ;    from  180°  to  270°,  it  increases  from  —  x  to 

—  1  ;  and  from  270°  to  360°,  it  decreases  from  —1  to  —  x. 

Note.  Wherever  the  symbol  oo  occurs  in  the  foregoing  discussion, 
it  must  be  interpreted  in  accordance  with  the  Note  to  Art.  41. 


42 


PLANE   TRIGONOMETKY. 


V.    GENERAL    FORMULAE. 

65.    To  fiyid  the  values  of  sin  {x  +  y)  and  cos  {x  +  y)  in 
terms  of  the  sines  and  cosines  of  x  and  y. 


C/ 


A 

k 

y 

/ 

/ 

B 

M^ 

-^^ 

/<» 

0 


A  .  D 


Let  AOB  and  BOC  denote  the  angles  x  and  ?/,  respec- 
tively ;  then,  AOC=x-{-y. 

From  any  point  C  in  OC  draw  QA  and  CB  perpendicular 
to  OA  and  OB ;  and  draw  BD  and  BE  perpendicular  to  OA 
and  AC. 

Since  EC  and  BC  are  perpendicular  to  0^  and  OB,  the 
angles  jBOjEJ  and  AOB  are  equal ;  that  is,  BCE  =  x. 

We  then  have 

.    ,     ,     .      AC     BD-^  CE     BD  ,  CE 


But 


and 


BD      BD  ^  OB 

= X =  sm  X  cos  y. 

OC      OB      OC  ^ 

CE      CE      BC 

— -  =  — -  X  — -  =  COS  a?  sinw. 
OC      BC      OC 


Whence,  sin  (x  -\-y)  =  sin  x  cos y  +  cos xsiny. 

.      .  ,     ,     .      OA      OD-BE      OD     BE 

Again,      cos(.  +  2/)=^=^^-  =  ^-^- 


(11) 


OD^OD      OB 
OC      OB      OC 


cosx  COS  2/, 


But 

and 

Whence,  cos (ic  -|-  ?/)  =  cos  x  cos  y  —  sin x  sin 2/. 


5^      BE      BC       . 

- —  = X  =  sni  X  sin  y. 

OC      BC      OC  ^ 


(12) 


GENERAL   FORMULA. 


43 


66.    To  find  the  values  of  sin  {x  —  y)  and  cos  {x  —  y)  in 
terms  of  the  sines  and  cosines  of  x  ojid  y. 


Let  AOB  and  BOC  denote  the  angles  x  and  ^,  respec- 
tively ;  then,  AOC  =  x  —  y. 

From  any  point  C  in  OC  di-aw  CA  and  CB  perpendicular 
to  OA  and  OB ;  also,  draw  BD  perpendicular  to  0^1,  and 
BE  perpendicular  to  AC  produced. 

Since  EC  and  BC  arc  perpendicular  to  OA  and  OB,  the 
angles  BCE  and  AOB  are  equal ;  that  is,  BCE  =  x. 

We  then  have 

sin(x'-y)  =  ^==^^^:^  =  ^-^. 
^  OC  OC  OC      00 


But 


and 


BD      BD  ^  OB 


CE     CE  ^  BC 

=  ^-T-.  X  — r  =  coso;  sin/y. 


OC     BC     OC 
Whence,  sin(x  —  y)=  sin  a;  cosy  —  cos  a;  sin?/. 

Again,      ,o.{x-y)  =  9A=.OD^BE^gD^BE^ 
^  ^       ^^      OC  OC  OC      OC 


(13) 


But 

and 


OD      OD  ^  OB 

oc^m'' oc= '''''' '""'y^ 

BE      BE     BC       . 


OC      BC      OC 
Whence,  cos  (a;  —  y)=  coso;  cos  2/  +  sin  ic  sin?/. 


(14) 


44 


PLANE   TRIGONOMETRY. 


67.  The  fundamental  formulae  of  Arts.  65  and  66  are  of 
great  importance,  and  it  is  necessary  to  prove  that  they  hold 
for  all  values  of  x  and  y. 

It  is  obvious  that  the  proof  of  Art.  65  is  not  general,  for 
we  have  assumed  in  the  construction  of  the  figure  that  x  and 
y  are  acute  angles,  and  that  x-\-y  is  <  90°.  Also,  in  Art. 
66,  we  have  taken  x  and  y  as  acute  angles,  and  x':>y. 

In  order  to  prove  the  formulae  universally,  we  will  first 
show  that  (ll)  and  (12)  hold  for  all  values  of  x  and  y,  and 
we  can  then  give  a  general  proof  of  (13)  and  (l4). 

68.  We  will  first  prove  (11)  and  (12)  when  x  and  y  are 
acute,  and  x-\-y>  90°. 


0    D 


Let  DOB  and  BOC  denote  the  angles  x  and  y^  respec- 
tively ;  then,  DOC  =  x  -\-y. 

From  any  point  O  in  0(7  draw  CB  perpendicular  to  OB, 
and  CA  perpendicular  to  OD  produced ;  and  draw  BD  and 
BE  perpendicular  to  OD  and  AC. 

Since  EC  and  BC  are  perpendicular  to  OD  and  OB,  the 
angles  BCE  and  DOB  are  equal ;  that  is,  BCE  =  x. 

We  then  have,  by  Art.  32, 


But 


and 


sin  DOC  = 

BD^ 
OC 

CE 


OC 


AC^BD-j-CE  ^BD      CE 

OC  OC  OC     OC 

BD      OB 

— —  X  ——  =  sm  X  cos  y, 
OB      OC  ^ 

CE      BC 

X  — -  =  COS  X  sm  y. 

BC      OC  ^ 


Whence,  sin(a^  -f-  ?/)  =  sinx  cosy  +  cosa?  sin 2/. 


Again,        cos  DOC  = 


GENERAL   FORMULAE.  45 

OA      OD-BE      OD     BE 


oc         oc         oc    oc 


OD      OD      OB 

But  —  = X =  cos  X  cos  v, 

OC      OB      OC  ^ 

,  BE     BE     BC       .         . 

and  = X  —  =  sin  x  sin  y. 

OC     BC      OC  ^ 

Whence,  cos(:(;  +  y)  =  cos  x  cos  y  —  sin  x  sin  y, 

69.  We  have  thus  proved  (ii)  and  (12)  when  x  and  y 
are  any  two  acute  angles;  or,  what  is  the  same  thing,  when 
they  are  any  two  angles  in  the  first  quadrant. 

Now  let  a  and  b  be  any  assigned  vaUies  of  x  and  y  for 
which  (11)  and  (12)  are  true  ;  then  by  Art.  44, 

sin  [90°+  (a +  6)]  =  cos  (a -h  6) 

=  cos  a  cos  6  — sin  a  sin  6,     by  (12);    (a) 
and, 

cos  [90°-f  (a -}-/>)]  =  -sin  («-h^) 

=  —  sin  a  cos  6  — cos  a  sin  Z>,  by  (11).     (b) 
But  by  Art.  44, 

cos  a  =  sin  (90°+  a) ,  cos  6  =  sin  (90°+  b) , 

—  sin  a  =  cos  (90°+ a),         —  sin  &  =  cos  (90°+ 6). 

Hence  (A)  may  be  written  in  the  forms, 

sin  [(90°+ a)  +6]  =  sin  (90°+ a)  cos  6  +  cos  (90°+ a)  sin  6, 
sin  [a  +  (90°+  b) ]  =  sin  a  cos  (90°+  b)  +  cos  a  sin  (90°+  b) , 

both  of  which  are  in  accordance  with  (11) . 
And  (b)  may  be  written  in  the  forms, 

cos  [(90°+  a)  +  6]  =  cos  (90°+  a)  cos  b-sin  (90°+  a)  sin  b, 
cos  [a+  (90°+ 6)]  =  cos  a  cos  (90°+ 6) -sin a  sin  (90°+ 6), 

both  of  which  are  in  accordance  with  (12). 

It  follows  from  the  above  that  if  (11)  and  (12)  hold  for 
any  assigned  values  of  x  and  y.,  such  as  a  and  6,  they  also 
hold  when  either  a  or  b  is  increased  by  90°. 


46  PLANE   TRIGONOMETRY. 

But  they  have  been  proved  to  hold  when  both  x  and  y  are 
in  the  first  quadrant ;  hence  they  also  hold  when  x  is  in  the 
second  quadrant  and  y  in  the  first.  And  since  they  hold 
when  X  is  in  the  second  quadrant  and  y  in  the  first,  they  also 
hold  when  x  is  in  the  thii-d  quadi'ant  and  y  in  the  first ;  and 
so  on. 

Thus  (ii)  and  (12)  are  proved  to  hold  for  any  values  of 
X  and  y  whatever,  positive  or  negative. 

70.  We  may  now  give  a  general  proof  of  (13)  and  (14). 
B3'  (11)  and  (12),  we  have 

sin  {x  -y)  =  sin  [a^'  +  (-  2/)] 

=  sin  x  cos  (  —  2/)  +  cos  x  sin  (  —  y) 
—  sin  X  cos  y  —  cos  x  sin  y  (Art.  42) . 
And,   cos  (x  —  y)==  cos  [a?  +  (  —  2/)  ] 

=  cos  X  cos(  —  y)  —  sin  x  sin(  —  y) 
=  cos  X  cos  ?/  +  sin  a;  sin  y  (Art.  42) . 

Hence  (13)  and  (14)  hold  for  all  values  of  x  and  y,  for 
the  above  proof  depends  on  formulae  which  have  been  shown 
to  hold  universally. 

71.  By  Art.  20,  we  have 

sin  {x  +  y) 

tan  (x  -f-  2/)  = 7 — ; — X 

^        ^^      cos{x  +  y) 

sin  X  cos  y  +  cos  a?  sin  w   ,      ^     ^ 

= ^-^ ;— ^,  by  (11)  and  (12) 

cos  X  COS  y  ~  sm  x  sin  y 

Dividing  each  term  of  the  fraction  by  cos  x  cos  y, 

sin  X  cos  y      cos  x  sin  y 

.       cos, X  cosy      cos 07  cos 2/ 

tan  (x  -\-y)  = : -. 

^   ^        cos  X  cos  2/   sm  X   sm  y 

cos  X  cos  y     cos  x  cos  y 

tan  X  +  tan  y  _  /  \ 

""  1  —  tan  X  tan  y 


GENERAL   FORMULAE.  47 

In  like  manner,  we  derive 

tan  a;- tan  y  ,     . 

tan  (X  -y)  =  i^^ana^tan^'  ^^^^ 

Again,  by  Art.  20,  we  have 

w     .     N      ^Qs  (x  +  y) 
cot  (a; -\-y)  =  —. — 7 — rz\ 

cos  .T  cos  2/ —  sin  a;  sin  2/   ,     ,     .      j /,^v 

= ^^-^  by  (11)  and  (12 )  • 

sin  X  cos  y  +  cos  x  sin  ?/     "  ^     ' 

Dividing  each  term  of  the  fraction  by  sin  x  sin  1/, 

cos  X  cos  2/      sin  a;  sin  y 

sin  X  sin  ?/      sin  x  sin  ?/ 
cot(a;H-?/)  =- 


sin  .T  cos  v      cos  ar  sin  y 
sm  X  sm  ?/      sin  x  sm  ?/ 

cot  X  cot  ?/  —  1 


(17) 


cot  y  +  cot  a; 

In  like  manner,  we  derive 

cot  a;  cot  y  -\- 1  ,     . 

^'      •-'/""    cot  ,y  —  cot  a; 

72.    From  (11),  (12),  (13),  and  (14),  we  obtain 
sin  (rt  -(-  />)  =  sin  a  cos  h  +  cos  a  sin  6, 
sin  (a  —  h)  —  sin  a  cos  6  —  cos  a  sin  Z>, 
cos  (a  H-  6)  =  cos  a  cos  h  —  sin  f/  sin  6, 
cos  (a  —  6)  =  cos  ft  cos  6  +  sin  a  sin  h. 

Whence,  by  addition  and  snbtraction, 

sin  (a  +  6)  +  sin  (ct  —  6)  =  2  sin  a  cos  &, 
sin  (ft  +  ?>)  —  sin  (fi  —  6)  =  2  cos  a  sin  6, 
cos  (a  +  &)  +  cos  (a  —  5)  =  2  cos  a  cos  5, 
cos  (ft  +  6)  —  cos  («  —  6)  =  —  2  sin  a  sin  6. 


48  PLANE   TRIGONOMETRY. 

Let  a  +  h=x,  and  a  —  h  =  y. 

Then,  a  =  |-(a;  +  2/) ,  and  5  =  ^(a;  —  2/) . 

Substituting  these  values,  we  have 

sin  X  +  sin  y  =  2  sin  ^{x  +  y)  cos  ^{x  —  y)^  (l9) 
sin  ft.'  —  sin  2/  =  2  cos  ^{x  +  2/)  sin  ^  (ft?  —  y) ,  (20) 
cosft;+cos?/=  2  cos^(ft;  +  ?/)  cos^(ft;  —  ?/),  (2l) 
cos  X  —  cos  2/  =  —2  sin  |-(ft;  +  y)  sin  ^(ft;  —  2/)  •        (22) 

73.  Dividing  (19)  b}^  (20),  we  obtain 

sin  X  +  sin  y  _  2  sin|-(a?  +  y)  cos  |- (ft.'  — 2/) 
sin  ft;  — sin  2/      2  cos  |- (ft; +  2/)  sin  ^  (ft;  — 2/) 

=  tan|-(ft;  +  ?/)  cot|(ft;-2/) 

^tan|-(ft;  +  2/)    (^^^t.  18).  (23) 

tan  ^  (ft;  — 2/) 

FUNCTIONS   OF  2  a:. 

74.  Putting  2/  =  ftJ  in  (11) ,  we  have 

sin  2  ft;  =  sin  x  cos  x  +  cos  x  sin  a; 

=  2  sin  ft;  cos  x.  (24) 

Putting  2/  =  a?  in  (12) , 

cos  2  ft;  =  cos^  X  —  sin^  ft;.  (25) 

Since  cos^  a?  =  1  —  sin^  ft;,  and  sin-  a;  =  1  —  cos^  ft?  (Art.  19) , 
we  also  have 

cos  2  ft;  =  1  —  sin^  x  —  sin^  x     =1  —  2  sin^  x,  (26) 

and  cos  2ft;  =  cos- ft;  —  (1  —  cos^ft;)=  2cos^ft;—  1.(27) 

In  like  manner,  from  (15)  and  (17) ,  we  have 

tan2ft;^     ^^^"^   ,  (28) 

1  —  tan-  X 

cot  2  ft;  =  ^^^-^^^.  (29) 

2  cot  X 


GENERAL  FORMULA.  49 

FUNCTIONS  OF  ^x. 

75.   From  (26)  and  (27),  we  have 

2  sin^ x=  1  —  cos  2 .i*,  and  2  cos- .i'  =  1  +  cos  2 x. 
Writing  x  in  place  of  2  x,  and  therefore  ^  a;  in  place  of  x, 

2  sin^  -^  a;  =  1  —  cos  x,  and  2  cos-  ^  .i-  =  1  +  cos  x.       (A ) 
Dividing  by  2,  and  extracting  the  square  root, 


8mix=  JLzl£21^ 


(30) 


C09  j-a;=    p  +cosa;  .  ^. 


2 

Dividing  (30)  b}^  (31),  we  ol)tain 


tania;=  jLz^2i^.  (32) 

\l  +  cosa;  ^  ^ 

Multiplying  the  terms  of  the  fraction  under  the  radical  sign 


first  by  1  +  cos  x,  and  then  by  1  —  cos  x,  we  have 
^x=  J  1-eos-a.- 


tan^a;  = 


(1  +  cos  a;)' 


-J 


sm^  X  sm  X 


(1  +  cos  a;)^      1  +  cos  x 


(33) 


and  tan  J  a;  =      (l-cosaQ^ 

\   1  —  cos-  a; 

_    1(1  —  cos  a;)^  _  1  —  cos  x 


(34) 


sin-  X  sm  X 

And  since  the  cotangent  is  the  reciprocal  of  the  tangent, 

^^+ 1  ^      1  +  cos  a;         sin  a;  ,     ^ 

cotia;  =  — ; :^- (35) 

sm  a;         1  —  cos  x 

Note.  The  radical  in  each  of  the  formulse  (30),  (31),  and  (32)  is 
to  be  taken  as  positive  or  negative  according  to  the  quadrant  in  which 
the  angle  ^a:  is  situated  (Art.  37). 


50  PLANE  TRIGONOMETRY. 

INVERSE  TRIGONOMETRIC  FUNCTIONS. 

76.  The  expression  sin"^^/?  called  the  inverse  sine  of  2/,  or 
the  anti-sine  of  y,  is  used  to  denote  the  angle  whose  sine  is 
equal  to  y. 

Thus  the  fact  that  the  sine  of  the  angle  x  is  equal  to  y  may 
be  expressed  in  either  of  the  waj^s 

sin  X  =  ?/,  or  x  =  sin~-^  y. 

In  like  manner,  the  expression  cos'^y  signifies  the  angle 
whose  cosine  is  equal  to  y  ;  tan~^  y,  the  angle  whose  tangent 
is  equal  to  ?/,  etc. 

Note.  The  student  must  be  careful  not  to  confuse  this  notation 
with  the  exponent  —1;  the  —1  power  of  sin  a:  is  expressed  (sinx)-^, 
and  not  sin-^a:. 

77.  By  aid  of  the  principles  of  Art.  76,  any  relation  in- 
volving du'ect  functions  may  be  transformed  into  one  involv- 
ing inverse  functions. 

Take  for  example  the  formula 

sin  (^x-\-y)  =  sin  x  cos  y  -\-  cos  x  sin  y.  (a) 

Let  sin  x  =  a,  and  sin  y  =  b. 

Then  by  Art.  76,  x=  sin~^  a,  and  y  =  sin~^  b. 


Also  by  Art.  19,     cos  x=  ^1  —  siia^x, 


and  cos  y  =  \1  —  sin^y. 

Putting  sin  x  =  a,  and  sin  y  =  b,  we  have 

cos  x=  ^1  —  a^,  and  cos  y=^l  —  b'. 
Substituting  these  values  in  (a)  , 

sin  (sin~^  a  +  sin"'^  b)  =  a^l  —  b^  -\-b^l  —  a^. 
Whence  by  Art.  76, 

sin^^  a  +  sin"^  b  =  sin~^  (o  \/l  —  6^  +  6  V^l  —  a^) . 


GENERAL  FORMULA.  53 

EXERCISES. 

1  tjiri  Tj 

78.   1.    Prove  the  relation  cos'2x= -^- 

1  +  tannic 

By  (25) ,       cos  '2x=  cos- it*  —  sin^x* 

cos-o;  —  sin-a;    ,.    ,    .f.. 

= :, — — — r-   (Art.  19). 

cos- a;  +  siQ-.^ 

Dividiug  each  term  of  the  fraction  by  cos-.i*,  we  have 

,        sin- X 

cos- a;      1  —  tan- a; 

cos  '2x  = 7-:r-  =  :; :; — 

1  -I-  ^^""  ^      1  H~  tan-a; 
cos- a; 

2-0          4.U       ^  4--        sin  5  a.' +  sin  a;      ,       „ 
.    Prove  the  relation    — — — ' =  tan  3  a;. 

cos  ox  -j-  cos X 
By  (19)  and  (21), 

sin  5 a;  +  sin  x _  2 sin  l(iJx-\-x)  cos ^(5 a;  — x) 
cos  5  a;  +  cos x      2  cos  ^ (5  a;  +  a*)  cos  ^ (5  a;  —  x) 

sin  3  a' 


cos  3  a; 

Prove  the  following  relations  : 

«     sin  (a;  -j-y)  _  tan  a-  -f  tan  y 
sin  (x  —  y)      tan  x  —  tan  y 


=  tan  3  X. 


4. 


cos  (a-  -f-  ?/)  _  cot  X  cot  //  —  1 


cos  (a;  —  y)      cot  x  cot  y  -\-l 

c     sin  (x  +  y)  _  1  -f  cot  X  tan  y 
cos  [x  —  y)        cot  X  -\-  tan  y 

c       '    /  4-0  ,     \       sin  y  -f-cos  y 

6.  sm(4o  +?/)  =  — ^— ^ ^. 

V^2 

7.  tan  (60° -3/)=    V^-tany. 

1  +  V^  tan  2/ 

Q     sin  aj  4-  sin  ?/      ,       ,  .     ,     . 

8.  — — ^  =  tani(a;H-?/). 

cos  a;  -|-  cos  y 


52  PLANE   TRIGONOMETRY. 

g_    sinx  +  sinj/=_coti(iC-^). 

COS  X  —  COS  y 

10.    sm2i»  = 


1  +  tan^aj 


11 .  CSC  2  a;  =  ^  sec  x  esc  a;. 

2 

12.  tan  a;  +  cot  a;  = 

sin  2  a? 

13.  cot  a?  —  tan  a;  =  2  cot  2  a;. 

14     (l  +  tana;)^-(l-tana;)^^^.^^^^ 
(l+tana?)2+  ^i  _tana;)2 

15.  sin  {x  +  y)  sin  {x  —  y)  =  siu^x  —  sin^i^. 

16.  cos  {x  +  y)  cos  (x  ~y)  =  cos- a;  —  sin^?/. 

17.  sec^a?  csc^a;  =  sec^a;  +  csc^a;. 

18.  cos  y  +  cos  (120°  -hy)  +  cos  (120°  -y)=0. 

19.  sin^sin(^~(7)  +  sin5sin((7- J.) 

+  sinOsin(^-5)  =  0. 

20.  cos  (A  +  B)  cos  {A-B)  +  cos (5  +  O)  cos  (5  -  O) 

+  cos(C+^)cos(a-^)  =  cos2J[+cos25+cos2(7. 

rti      cos  a?  —  cos  ox      .      ^ 

21 .    =  tan  2  a;. 

sin  3  a?  —  sin  x 

22     cos  80°  +  cos  20°  _  ^/^ 
sin  80°  -  sin  20° 

By  putting  x  =  x-]-y  and  y  =  z  in  (ii)  and  (12) ,  Art.  65, 
prove : 

23.  sin  (x  -\-y  -\-z)  =  sin  a?  cosy  cosz  +  cosa?  sin  2/  cos;^; 

+  cos  a?  cosy  sinz  —  sina?  siny  sinz. 

24.  cos  {x-\-y  -{-z)  =  cos  x  cos  ?/  cos  z  —  sin  a?  sin  y  cos  2; 

—  sin  a?  cosy  sinz  —  cos  a;  sin  2/  sins;. 


GENERAL   FORMULAE.  53 

By  putting   aj=2a;   and   y  =  x   in  (ll),  (12),  and  (15), 
prove  : 

25.  sin  3  a;  =  3  sin a^  —  4  sin^a;. 

26.  cos3i«  =  4cos^a;  —  3  cosic. 

oty     4.      o         3  tana;  — tan^a; 

27.  tan  3  x  = 

1  -3tan2a; 
Prove  the  relations : 

28.  sin  {2x-\-y)  —  2  sin x  cos  {x  -\-y)  =  siny. 

QQ     sin  3  a;  _  cos  3  a;  _  ,-^ 
sin  a;        cos  a; 

30.  1  4- cos  2 a; cos  2y=2  (sin- .r  sin-?/  +  cos^ajcos-?/). 

31 .  1  -h  tan  X  tan  2  x  =  sec  2  a;. 

32.  sin  4  .t  =  4  sin  a;  cos  x  —  S  s'rn^x  cos  x. 

33 .  cos  4  a;  =  8  cos^  x  —  8  cos-  x  +  1 . 

34.  sin  5  a;  =  5  sin  a;  —  20  siu'^v  +16  sin' a;. 

35.  By  putting  a;=  45°  and  y  =  SO°  in  (13)  and  (14),  Art. 

66,  prove 

sin  15°  =  ^  (v/6  -  v/2)  =  cos  75°, 
cos  15°  =  ^  (\/6  4-  V^)  =  sin  75°.      ' 

36.  By  putting  .^' =  30°  in  (34)  and  (35),  Art.  75,  prove 

tanl5°  =  2-V^  =  cot75°, 

cot  15°  =  2  +  V^  =  tan75°. 

37.  By  putting  a;  =  45°  in  (30)  and  (31),  Art.  75,  prove 

sin22°30'  =  iV/2-V^,  cos  22°30' =  i\/2 -j- v/2. 

38.  By  putting  a;  =  45°  in  (34)  and  (35),  Art.  75,  prove 

tan22°30'  =  V^-l,    cot 22°30'  =  V^  +  1. 


54  PLANE  TRIGONOMETRY. 


VI.    LOGARITHMS. 

79.  Every  positive  number  may  be  expressed,  exactly  or 
approximately,  as  a  power  of  10  ;  thus, 

100=102;  13  =  10i"39-- ;  etc. 

When  thus  expressed,  the  corresponding  exponent  is 
called  its  Logarithm  to  the  base  10  ;  thus,  2  is  the  logarithm 
of  100  to  the  base  10,  a  relation  which  is  written 

logio  100  =  2,  or  simply  log  100  =  2. 

And  in  general,  if  10"=  =  m,  then  x  =  log  m. 

80.  Any  positive  number  except  unity  may  be  taken  as 
the  base  of  a  system  of  logarithms;  thus,  if  a^'^m,  then 
a;  =  log^m. 

Logarithms  to  the  base  10  are  called  Common  Logarithms, 
and  are  the  only  ones  used  in  numerical  computation. 
If  no  base  is  expressed,  the  base  10  is  understood. 

81.  We  have  by  Algebra, 

10°=  1,  10-'  =  iiff     =.1, 

101  =  10,  10-'  =  Ti7    =.01. 

102=100,  10-3  =  y^=. 001,  etc. 

Whence,  by  the  definition  of  Art.  79, 

log  1  =  0,  log.l  = -1  =  9-10, 

log  10=1,  log  .01=  -2  =  8-10, 

log  100  =  2,  log  .001  =  -3  =  7  -  10,  etc. 

Note.  The  second  form  of  the  results  for  log.l,  log  .01,  etc.,  is 
preferable  in  practice. 

82.  It  is  evident  from  Art.  81  that  the  logarithm  of  a 
number  greater  than  1  is  positive,  and  that  the  logarithm  of 
a  number  between  0  and  1  is  negative. 


LOGARITHMS.  55 

83.  If  a  number  is  not  an  exact  power  of  10,  its  common 
logarithm  can  only  be  expressed  approximately  ;  tiie  integral 
part  of  the  logarithm  is  called  the  characteristic^  and  the 
decimal  part  the  mantissa. 

For  example,  log  13  =  1.1139. 

In  this  case  the  characteristic  is  1,  and  the  mantissa  .1139. 

84.  It  is  evident  from  the  first  column  of  Art.  81  that  the 
logarithm  of  any  number  between 

1  and      10  is  equal  to  0  plus  a  decimal ; 
10  and    100  is  equal  to  1  plus  a  decimal ; 
100  and  1000  is  equal  to  2  plus  a  decimal ;  etc. 

Hence,  the  characteristic  of  the  logarithm  of  a  number 
with  one  figure  to  the  left  of  its  decimal  point,  is  0  ;  with 
two  figures  to  the  left  of  the  decimal  point,  is  1  ;  with  tliree 
figures  to  the  left  of  the  decimal  point,  is  2  ;  etc. 

85.  In  like  manner,  from  the  second  column  of  Art.  81, 
the  logarithm  of  a  decimal  between 

1  and      .1  is  equal  to  9  plus  a  decimal  —  10  ; 
.1  and    .01  is  equal  to  8  plus  a  decimal  —  10  ; 
.01  and  .001  is  equal  to  7  plus  a  decimal  —  10  ;  etc. 

Hence,  the  characteristic  of  the  logarithm  of  a  decimal 
with  no  ciphers  between  its  decimal  point  and  first  significant 
figure,  is  9,  with  —10  after  the  mantissa;  of  a  decimal  with 
one  cipher  between  its  point  and  first  figure,  is  8,  with  —10 
after  the  mantissa ;  of  a  decimal  with  two  ciphers  between 
its  point  and  first  figure,  is  7,  with  —10  after  the  mantissa ; 
etc. 

86.  For  reasons  which  will  be  given  hereafter,  only  the 
mantissa  of  the  logarithm  is  given  in  tables  of  logarithms  of 
numbers  ;  the  characteristic  must  be  supplied  by  the  reader. 

The  rules  for  characteristic  are  based  on  Arts.  84  and  85  : 


56  PLANE   TRIGONOMETRY. 

I.    If  the  number  is  greater  than  1,  the  characteristic  is  1 
less  than  the  number  of  places  to  the  left  of  the  decimal  point. 

II.  If  the  number  is  between  0  and  1,  subtract  the  number 
of  cip>hers  betiveen  the  decimal  point  and  first  significant  figure 
from  9,  ivriting  —10  after  the  mantissa. 

Thus,  characteristic  of  log  906328.5  =  5  ; 

characteristic  of  log  .007023    =7,  with  —10  after 
the  mantissa. 

Note.  Some  writers,  in  dealing  with  the  characteristics  of  negative 
logarithms,  combine  the  two  portions  of  the  characteristic,  writing  the 
result  as  a  negative  characteristic  before  the  mantissa. 

Thus,  instead  of  7.6036  —  10,  the  student  will  frequently  find  3^.6036, 
a  minus  sign  being  written  over  the  characteristic  to  denote  that  it 
alone  is  negative,  the  mantissa  being  always  positive. 

PROPERTIES  OF  LOGARITHMS. 

87.  In  any  system.,  the  logarithm  of  unity  is  zero. 
For  since  a°  =  1,  we  have  log^  1  =  0  (Art.  79). 

88.  In  any  system,  the  logarithm  of  the  base  itself  is  unity. 
For  since  a^  =  a,  we  have  log„  a  =  1 . 

89.  In  cmy  system  ivhose  base  is  greater  than  unity,  the 

logarithm  of  zero  is  minus  infinity. 

For  if  a  is  >  1 ,  we  have  (x~°°  =  —  =  —  =  0. 
Whence  by  Art.  79,  log^  0  =  —  oo  . 

90.  In  any  system,  the  logarithm  of  a  product  is  equal  to 
the  sum  of  the  logarithms  of  its  factors. 

Assume  the  equations 

a""  =  m^         ,  ,  c  x  =  loff„ m, 

[  ;  whence  by  Art.  79,  ]  ^"     ' 

a^  =  n  }  "  (y  =ilogan. 


LOGARITHMS.  57 

Multiplying,  we  have 

a""  X  a^  =  mn^  or  a"""^^  =  mn. 
Whence,         log^  7nn  =  x  -\-y. 
Substituting  the  values  of  x  and  y,  we  have 

log,,  mn  =  log,,  m  ~\-  log„  n . 

In  like  manner,  the  theorem  may  be  proved  for  the  product 
of  three  or  more  factors. 

91.  By  aid  of  the  theorem  of  Art.  90,  the  logarithm  of 
an}'  composite  number  may  be  found  when  the  logarithms  of 
its  factors  are  known. 

1.    Given  log  2  =  .3010,  log  3=  .4771  ;  find  log  72. 

log  72  =  log  (2x2x2x3x3) 

=  log  2  -f  log  2  -h  log  2  +  log  3  +  log  3 
=  3  X  log  2  -h  2  X  log  3 
=  . 9030 +  .!)5-i2  =  1.8572. 

EXAMPLES. 

Given  log2  =  .3010,  log  3  =  .4771,  log  5  =  .0990,  logT 
=  .8451  ;  find  the  values  of  the  following : 

2.  log  6.  7.  log  21.  12.  log  98.  17.  log  135. 

3.  log  14.  8.  log  63.  13.  log  105.  18.  log  168. 

4.  log 8.  9.  log 56.  14.  log  112.  19.  log  147. 

5.  log  12.  10.  log  84.  15.  log  144.  20.  log 375. 

6.  log  15.  11.  log 45.  16.  log  216.  21.  log  343. 

92.  In  any  system.,  the  logaritlim  of  a  fraction  is  equal  to 
the  logarithm  of  the  numerator  minus  the  logarithm  of  the 
denominator. 

Assume  the  equations 

a""  =  m]  (  x  =  \oga  m, 


>■  ;  whence. 


a'-"  =  n  )  (y  =  logrt'^i- 


58  PLANE   TRIGONOMETRY. 


Dividing,  we  nave   _  =  — ,    or    cC^  =  — 

cO'      n  n 

Whence,  log„  —  =  x  —  y  =  log,,  m  —  log„  n, 

93.    1.    Griven  log  2  =  .3010  ;  find  log  5. 

log  5  =  log  i5  =  log  10  -  log  2  =  1  -  .3010  =  .6990. 


EXAMPLES. 

Given  log  2  =  .3010,  log  3  =  .4771,  log  7  =  .8451  ;  find  the 
values  of  the  foUowins: : 


2. 

lopj  — 
^3 

5. 

log  35. 

8. 

1      42 

loo- 

"25 

11. 

log  71. 

3. 

1      10 

logy. 

6. 

1      21 
log  — 
=^16 

9. 

log  175. 

12. 

1      35 
loar  — 
°  6 

4. 

log  31 

7. 

log  125. 

10. 

log  111. 

13. 

log5f 

94.  In  any  system,  the  logarithm  of  any  poiver  of  a  quan- 
tity is  equal  to  the  logarithm  of  the  quantity  multiplied  by  the 
exponent  of  the  poioer. 

Assume  the  equation 

cf  =  m  ;  whence,  x  =  log^  m. 

Raising  both  members  to  the  j?th  power,  we  have 

a^""  =  m^  ;  whence,  log„  m^  =px  =p  log„  m. 

95.  In  any  system,  the  logarithm  of  any  root  of  a  qnantity 
is  equal  to  the  logarithm  of  the  quantity  divided  by  the  index 
of  the  root. 

r,—  I  1 

For,         log„  V m  =  log,  (m')  =  -  log„  m  (Art.  94) . 

96.  1.    Given   log  2  =  .3010;  find  the  logarithm  of  2i 

log  23  =  5  X  log  2  =  -  X  .3010  =  .5017. 

Note.  To  multiply  a  logarithm  by  a  fraction,  multiply  first  by  the 
numerator,  and  divide  the  result  by  the  denominator. 


LOGARITHMS.  59 

2.    Given  log  3  =  .4771  ;  find  the  logarithm  of  V^S. 

,      Ao      loo"  3       .4771        ,^xap 
log  y o  =     "    =  — —  =  .Uoyb. 


EXAMPLES. 

Given  log  2  =  .3010,  log  3  =  .4771 ,  log  7  =  .8451  ;  find  the 
values  of  the  follow  in  oj : 


3. 

logSi 

7. 

log  12'. 

11. 

log  IS*^. 

15. 

6/- 
logyo. 

4. 

log  2^ 

8. 

log  21^. 

12. 

logN/7. 

16. 

log  y35. 

5. 

log7^ 

9. 

logl4^ 

13. 

log  fd. 

17. 

log  y'98. 

6. 

log  5^. 

10. 

log25i 

14. 

logV'2. 

18. 

logV^126. 

19.    Find  tlio  logarithra  of  (2^  x  3'') . 

By  Art.  90,  log  (2'  x  3 ')=  log  2'  +  log  s' 

=  Jl()g2+ Jlog3 
=  .1003 +  .5964  =.6967. 

Find  the  values  of  the  following  : 

20.  logl'yj.  22.  log(3'  X2-).  24.  \ogyjl    26.  logy/y- 

21.  log-.        23.  log3y^7.  25.  logfl.    27.  log*^. 

97.    In  the  common  system,  the  mantissce  of  the  logarithms 
of  numhers  having  the  same  sequence  of  figures  are  equal. 

To  illustrate,  suppose  that  log  3.053  =  .4847. 
Then, 

log  30.53    =  log  (10  X  3.053)    =  log  10  +  log  3.053 
^l  +  .4847  =1.4847; 

log  305.3    =  log  (100  X  3.053)  =  log  100  -h  log  3.053 
=  2 +  .4847  =2.4847; 

log  .03053  =  log  (.01  X  3. 053)  =  log. 01  +  log  3. 053 
=  8  -  10  +  .4847      =  8.4847  -  10  ;  etc. 


60  PLANE   TRIGONOMETRY. 

It  is  evident  from  the  foregoing  that  if  a  number  is  multi- 
plied or  divided  by  any  integral  power  of  10,  thus  produc- 
ing another  number  with  the  same  sequence  of  figures,  the 
mantissas  of  theii'  logarithms  will  be  equal. 

Thus,  if  log  3.053  =  .4847,  then 

log  30.53  =  1.4847,  log  .3053  =  9.4847  -  10, 
log  305.3  =  2.4847,  log  .03053  =8.4847-10, 
log  3053.  =  3.4847,         log  .003053  =  7.4847  -  10,  etc. 

Note.  The  reason  will  now  be  seen  for  the  statement  made  in 
Art.  86,  that  only  the  mantissas  are  given  in  a  table  of  logarithms  of 
numbers.  For,  to  find  the  logarithm  of  any  number,  we  have  only  to 
take  from  the  table  the  mantissa  corresponding  to  its  sequence  of 
figures,  and  the  characteristic  may  then  be  prefixed  in  accordance  with 
the  rules  of  Art.  86. 

This  property  of  logarithms  is  only  enjoyed  by  the  common  system, 
and  constitutes  its  superiority  over  others  for  the  purposes  of  numeri- 
cal computation, 

98.   1.  Given  log 2=. 3010,  log 3=. 4771;  find  log  .00432. 
log  432  =  log  (2*  X  3^)  =  4  log  2  +  3  log  3 
=  1.2040  4- 1.4313  =  2.6353. 
Then  by  Art.  97,  the  mantissa  of  the  result  is  .6353. 
Whence  by  Art.  86,      log  .00432  =  7.6353  -  10. 

EXAMPLES. 

Given  log  2  =  .3010,  log  3  =  .4771,  log  7  =  .8451  ;  find 
the  values  of  the  following : 

2.  log  1.8.  7.  log  .0054.  12.  log  302.4. 

3.  log  2.25.  8.  log  .000315.  13.  log  .06174. 

4.  log  .196.  9.  log  7350.  14.  log  (8.1)^ 

5.  log  .048.  10.  log  4.05.  15.  log  \/ 9^. 

6.  log  38.4.  11.  log  .448.  16.  log  (22.4)i 


LOGARITHMS.  61 

99.    To  prove  the  relation 

Assume  the  equations 

ce=m)  f  a.' =  log,,  771, 

\  ;    whence,  < 
h"  =  m  )  (.  V  =  log,,  771. 

From  the  assumed  equations,  we  have 

X 

cv'=b^,  or  aJ'  =  b. 
Whence,  log„6  =  '^    or  y 


y  loga& 

Substituting  the  values  of  x  and  y, 

loo"   777, 
log.  7?l  =  ,  ,  • 

log„6 

By  aid  of  this  relation,  if  the  logarithm  of  a  quantity  m  to 
a  certain  base  a  is  known,  its  logaritlnu  to  any  otlier  base  b 
may  be  found  by  dividing  by  the  logarithm  of  b  to  the  base  a. 

100.  To  jyrove  the  relation 

logj a  X  log,,  b=  I. 

Putting  7?7  =  a  in  the  result  of  Art.  1)1),  we  have 

log,,  a  1 

log,a  =  j^  =  . (Art.  88) . 

log,  6      \ogJj    ^  ^ 

Whence,  log,, a  x  log,, b=  \. 

APPLICATIONS. 

101.  The  value  of  an  arithmetical  quantity,  in  which  the 
operations  indicated  involve  only  nuiltiplication,  division, 
involution,  or  evolution,  may  be  most  conveniently  found  by 
logarithms. 

The  utility  of  the  process  consists  in  the  fact  that  addition 
takes  the  place  of  multiplication,  subtraction  of  division, 
multiplication  of  involution,  and  division  of  evolution. 

In  operations  with  negative  characteristics  the  rules  of 
Algebra  must  be  followed. 


62  PLANE   TRIGONOMETRY. 

102.     1.  Find  the  value  of  .0631  x  7.208  x  .51272. 

By  Art.  90,      log  (.0631  x  7.208  x  .51272) 

=  log  .0631  +  log  7.208  +  log  .51272. 

log  .0631    =    8.8000-10 

log  7.208    =    0.8578 

log  .51272=    9.7099-10 
Adding,    .-.  log  of  result  =  19.3677  -  20 

=    9.3677-10   (see  Note  1) 
Number  corresponding  to  9.3677—10  =  .23317. 

Note  1.    If  the  sum  is  a  negative  logarithm,  it  should  be  reduced  so 
that  the  negative  portion  of  the  characteristic  may  be  — 10. 
Thus,  19.3677-20  is  reduced  to  9.3677-10. 

2.    Find  the  value  of ' — 

7984 

By  Art.  92,  log  ^?^  =  log  336.8  -  log  7984. 

log  336.8  =  12.5273  -  10  (see  Note  2) 
los7984  =    3.9022 


Subtracting,  .-.  log  of  result  ='8.6251-10 
Number  corresponding  =  .04218. 

Note  2.  To  subtract  a  greater  logarithm  from  a  less,  or  to  subtract 
a  negative  logarithm  from  a  positive,  increase  the  characteristic  of  the 
minuend  by  10,  writing  —10  after  the  mantissa  to  compensate. 

Thus,  to  subtract  3.9022  from  2.5273,  write  the  minuend  in  the  form 
12.5273  -  10  ;  subtracting  3.9022  from  this,  the  result  is  8.6251  -  10. 

3.    Find  the  value  of  (.07396)^ 

By  Art.  94,  log  (.07396)^  =  5  x  log  .07396. 
log  .07396  =  8.8690 -10 

5 

44.3450  -  50 
=  4.3450  -  10  (see  Note  1) 
=  log  .000002213. 


LOGARITHMS.  63 


4.    Find  the  value  of  y. 035063. 

By  Art.  95,    log V^.035063  =  ^ log. 035063. 
log.035063  =  8.5440 -10 

20.  -  20  (see  Note  3) 

3)28.5449  -~30 

9.5150-  10  =  log  .3274. 

Note  3.  To  divide  a  negative  logarithm,  add  to  botli  parts  such  a 
multiple  of  10  as  will  make  the  negative  portion  of  the  characteristic 
exactly  divisible  by  the  divisor,  with  —10  as  the  quotient. 

Thus,  to  divide  8.5440  —  10  by  3,  add  20  to  both  parts  of  the  loga- 
rithm, giving  the  result  28.5449  —  30.  Dividing  this  by  3,  the  quotient 
is  9.5150-10. 

ARITHMETICAL  COMPLEMENT. 

103.  The  Arithmetical  Complement  of  the  logarithm  of  a 
number,  or  briefly  the  Colorjarithm  of  the  number,  is  the  loga- 
rithm of  the  reciprocal  of  that  number. 

Thus,  color?  409  =  lojr  --  =  locr  1  _  W  409. 

"^  ^400  ^  =' 

log  1  =  10.  -  10   (Note  2,  Art.  102) 

log  409  =.  2.G117 
.-.  colog409=    7.3883-10. 

Again,       colog  .067  =  log =  \o^  1  —  log  .067. 

.067 

log  1  =  10.  -10 

log  .067=    8.8261  -10 
.-.colog  .067=    1.1739. 
The  following  rule  is  evident  from  the  above  : 

To  find  the  cologarithm  of  a  number^  subtract  its  logarithm 
from  10  -  10. 

Note.  The  cologarithm  may  be  obtained  from  the  logarithm  by 
subtracting  the  last  significant  figure  from  10  and  each  of  the  others 
from  9,  — 10  being  written  after  the  result  in  the  case  of  a  positive 
logarithm. 


64  PLANE   TRIGONOMETRY. 

.51384 


104.    Example.    Find  the  valiie  of  — :: 


8.709  X  .0946 


,  .51384  .       f  K^oo^  sy      1      ^  _J_^ 

I02: =  log    .51o84  X X  

^  8.709  X  .0946  ^  ^  ^-709      .0946; 


=  log  .51384  +  log  -^ h  log 


8.709         ^  .094(; 
=  log  .51384+colog  8.709  +  colog. 0946. 

log  .51384  =  9.7109 -10 
colog  8.709  =  9.0601  -  10 
colog  .0946  =  1.0241 

9.7951 -10  =  log  .6239. 

It  is  evident  from  the  above  that  the  logarithm  of  a  fraction 
is  equal  to  the  logarithm  of  the  numerator  plus  the  cologa- 
rithm  of  the  denominator. 

Or  in  general,  to  find  the  logarithm  of  a  fraction  whose 
terms  are  composed  of  factors, 

Add  together  the  logarithms  of  the  factors  of  the  numerator, 
and  the  cologarithms  of  the  factors  of  the  denominator. 

Note.  The  value  of  the  above  fraction  may  be  found  without  using 
cologarithms,  by  the  following  formula  : 

log -^^^ =  log  .51384  -  log  (8.709  X  .0946) 

"=  8.709  X  .0946 

=  log  .51384  -  (log  8.709  +  log  .0946). 

The  advantage  in  the  use  of  cologarithms  is  that  the  written  work 
of  computation  is  exhibited  in  a  more  compact  form. 

EXAMPLES. 

105.  Note.  A  negative  quantity  can  have  no  common  logarithm, 
as  is  evident  from  the  definition  of  Art.  79.  If  negative  quantities 
occur  in  computation,  they  may  be  treated  as  if  they  were  positive,  and 
the  sign  of  the  result  determined  irrespective  of  the  logarithmic  work. 

Thus,  in  Ex.  3,  p.  65,  the  value  of  721.3  X  (-3.0528)  may  be  ob- 
tained by  finding  the  value  of  721.3  X  3.0528,  and  putting  a  negative 
sign  before  the  result.     See  also  Ex.  34,  p.  QQ. 


LOGARITHMS.  65 

Find  by  logarithms  the  vaUies  of  the  following : 

1.  9. 238  X. 9152.  4.    (-4.3264)  x(-. 050377). 

2.  130.36  X  .08237.  5.    .27031  X  .0-42809. 

3.  721.3  X  (-3.0528).  6.  ^(-.063165)  x  11.134. 

52.37*  '     .0.SG59  '  '    64327* 

8     LM?!.  10        9-163  ^2       -007514 


10.813  .0051422  -.015822 

13  3.3681  ^^     15.008  x(-.0843) 

12.853  X  .6349*  '        .06376  X  4.248 


15. 
16. 


(-2563)  X. 03442 
714.8x(-.511)  * 

121.6  X  (-9.025) 


(_48.3)  x3662x  (-.0856) 

17.  (23.86)\  22.    (.8)'.  28.    v/.4294. 

18.  (.532)«.  23.    (-3.16)3.         29.    v'.^2305. 

19.  (-1.0246)'.      24.    (.021)'.  30.    V^IOOO. 

20.  (.09323)^  25.    v/2.  31.    V^-. 00951. 

21.  5i  26.    v^5.  32.    V^.OOOlOll. 


33.    Find  the  value  of 


27.   V-3. 
2V^ 


3t 


log ?V^  =  log  2  +  log  ^5  +  colog  3^ 

=  log2  +  |log  5  +  f  colog  3. 

log  2=   .3010 
log  5=    .6990;  divide  by  3=    .2330 

colog  3  =  9.5229-10  ;  multiply  by  |= 9. 6024  -  10 


.1364  =  logl.3691. 


56  PLANE   TRIGONOMETKY. 


34.    Find  the  value  of  Ij — '  f~     ■ 

7.962 


logt^55296^ai     :03296_^j       Q329^ 
^  V  7.962       ^     ^  7.962        ^^    ^  &  ^ 

log  .03296=    8.5180-10 

log  7.962    =    0.9010 


3)27.6170-30 

9.2057- 10  =  log. 16059.     - 

Kesult,  -.16059. 


Find  the  values  of  the  following 


35.    2i-x3i  40.    r^^^^y.  45.    (^1^?^. 

\. 1321  J  \49309 

qfi     3^  ,-  46     ^-31-63Vt 

^^'  3*         41.  d-  ^"^ 


13 


3 


37. 


5^  , 47 


1003 


(_10)t*  42.    f^.  (.7325)^ 


f^  „         „  48.    ^^^Qg^. 

[j)'  43.    ^^^^1  S/.0008276 

39.    r^¥.  44.    ^x^3x^:0T.      49.    (-7469)^. 


38. 


113y  -(.2345)^ 

Krt        V''."0073  52.    (538.2x.0005969)i 

OU.    J' 

(•68291)^  53.    (18.9503)iix(-.l)'^ 


51. 


V/5.955XV61.2  54.    V3734.9  x  .00001108. 

V^298.54  55.    (2.6317)*  X  (.71272)^ 


gg     V -. 008193  x(.06285)t 
-  .98342 


57.    V^035x  y. 62667  X  y. 0072103. 


SOLUTION  OF  RIGHT   TRIANGLES. 


67 


VII.    SOLUTION   OF  RIGHT  TRIANGLES. 

106.  The  six  elements  of  a  triangle  are  its  three  sides  and 
its  three  angles. 

We  know  by  Geometry  that,  in  general,  a  triangle  is  com- 
pletely determined  when  three  of  its  elements  are  known, 
provided  one  of  them  is  a  side.  The  solution  of  a  triangle 
is  the  process  of  computing  the  unknown  from  the  given 
elements. 

107.  To  solve  a  rigJit  triangle,  tico  elements  must  be  given 
in  addition  to  the  right  angle,  one  of  which  must  be  a  side. 

The  various  cases  which  can  occur  may  all  be  solved  by 
aid  of  the  following  formula? : 


sin  A  = 

sin  B  = 


cos  A=  f 

c 


cos  B  = 


A  ^ 

tan^  =  Ti 

0 


tan  B  = 


a 


Case  I.    When  the  given  elements  are  a  side  and  an  angle. 

The  proper  formula  for  computing  either  of  the  remaining 
sides  may  be  found  by  the  following  rule : 

Take  that  function  of  the  angle  which  involves  the  given  side 
and  the  required  side. 

Case  II,    When  both  the  given  elements  are  sides. 

First  calculate  one  of  the  an2;les  bv  aid  of  either  of  the 
formulae  involving  the  given  elements,  and  then  compute  the 
remaining  side  as  in  Case  I. 


68  PLANE   TRIGONOMETRY. 

EXAMPLES. 

108.    1.  Given  c=  203.76,  5=21°  43'.     Find  a  and  6. 
In  this  case  the  formulae  to  be  used  are 

cos  jB  =  -9    and   sin  5  =  -  • 

c  c 

AVhence,  a-=  c  cos  jB,  and  h  =  c  sin  B. 

By  logarithms,     log  a  =  log  c  +  log  cos  5, 
and  log  h  =  log  c  +  log  sin  B. 

log  c  =  2.3091  log  c  =  2.3091 

log  cos  B  =  9.9681  -  10         log  sin  B  =  9.5682-10 

log  a  =  2.2772  log  b=  1.8773 

.-.  a=  189.3.  .'.  6=75.38. 

2.  Given  a  =  13,  A  =  67°  7'.     Find  6  and  c. 

.a         -,     •      ^      ^ 
In  this  case,  tan  A  =  ti   and  sm  A  =  -' 

'  6  c 

Whence,  6  = ^,    andc=^ — -. 

tan  A  sm  A 

By  logarithms,         log  6  =  log  a  —  log  tan  A, 

and  log  c  =  log  a  —  log  sin  A. 

log  a  =1.1139  log  a  =1.1139 

log  tan  A  =  0.3746  log  sin  A  =  9.9644-10 

log  b  =  0.7393  log  c  =  1.1495 

.-.  &  =  5.486.  .-.  c=  14.11. 

3.  Given  6=  .1512,  c=. 3081.     Find  ^  and  a. 

b  a 

In  this  case,    cos  A  =  -■>    and  tan  A  =  t^  or  a=b  tan  A. 

c  b 

Whence,    log  cos  A  =  log  b  —  log  c, 

and  log  a  =  log  b  -\-  log  tan  ^. 


SOLUTION   OF   RIGHT   TRIANGLES.  69 

loo-  h  =  9.1796  log  h  =  9.1796 

log  c  =  9 . 488 7  log  tau  A  =  0.2493 

log  cos  A  =  9.6909  log  a  =  9.4289 

.-.  J.  =  60°  36.4'.  .-.  a=    .2685. 

Note.  It  is  customary  in  practice  to  omit  writing  the  — 10  after  the 
mantissa  of  a  negative  logarithm,  as  illustrated  in  Ex.  3. 

109.  In  the  Trigonometrical  solution  of  a  triangle  bj'  the 
method  of  Case  II.,  it  is  necessary  to  first  find  one  of  the 
angles,  and  the  remaining  side  may  then  be  calculated. 

It  is  possible  however  to  obtain  the  third  side  directly, 
without  first  finding  the  angle,  by  Geometrical  methods. 

Thus  in  Ex.  3,  Art.  108,  w^e  have  by  Geometry, 

a-  -\-b~  =  c". 


Whence,  a  =  ^r -b- =  \/{c -{-b)  (c-b). 

By  logarithms,  log  a  ==  \  [log  (c  -|-  6)  -+-  log  (c  —  6)] . 

c-}-b=  .4593  ;  log  =9.6621-10 

c^b=  .1569;  log  =9.1956-10 
2)18.8577-20 
log  a  =  9.4289 -10 
.*.  a=    .2685,  as  before. 

If  the  given  sides  are  a  and  b,  the  formula  for  c  is 
y^a^  -f-  b^t  which  is  not  adapted  to  logarithmic  computation. 

In  such  a  case  it  is  usually  shorter  to  proceed  according 
to  the  rule  of  Art.  107. 

EXAMPLES. 
110.    Solve  the  following  right  triangles  : 

1.  Given  ^  =  43°  30',  c=11.2. 

2.  Given  B  =  68°  50',  a  =  729.3. 

3.  Given  i5=  62°  56',  6  =  47.7. 

4.  Given   a  =.624,  c=.91. 


70  PLANE   TRIGONOMETRY. 

5.  Given  a  =5,  6=2. 

6.  Given  ^  =  72°  7',  a  =  83.4. 

7.  Given  B  =  32°  10',  c  =  .02728. 

8.  Given   6  =  2.887,  c  =  5.11. 

9.  Given  ^  =  52°  41',  6  =  4247. 

10.  Given  a  =101,  6  =  116. 

11.  Given  ^  =  43°  22',  o  =  158.3. 

12.  Given  ^  =  58°  39',  c  =  35.73. 

13.  Given   a  =204.2,  c=  275.3. 

14.  Given  5  =  30°,  6  =  1.6438. 

15.  Given  ^=22°  14',  6=13.242. 

16.  Given  5=10°  51',  c=.7264. 

17.  Given   a  =  638.5,  6  =  501.2. 

18.  Given  ^=78°  17',  a  =  203.8. 

19.  Given   6  =.02497,  c=. 04792. 

20.  Given  5=2°  19'  30",  a  =  1875.3. 

21.  Given   a  =24.67,  6  =  33.02. 

22.  Given   6=1.4367,  c  =  3.4653. 

23.  Given  5=6°  12.3',  c  =  37206. 

24.  Given  ^=64°  1.3',  6  =  200.05. 

25.  Given   a  =  340.06,  6  =  231.69. 

26.  Given  a  =1.7087,  c  =  2.0008. 

27.  Given  5=21^33' 51",  a  =.82109. 

28.  Given  ^=  74°  0' 18",  c  =  275.62. 

29.  Given  5=34°  14' 37",  6  =  120.22. 

30.  Given   a  =  10.107,  6  =  17.303. 


SOLUTION   OF   RIGHT   TRIANGLES.  71 

Solve  the  followiog  isosceles  triangles,  in  which  A  and  B 
are  the  equal  angles,  and  o,  6,  and  c  denote  the  sides  oppo- 
site the  angles  A^  jB,  and  C,  respectively : 

31.  Given  ^1=68°  57',  Z>  =  35.091. 

32.  Given  5  =27°  8',  c  =  3.088. 

33.  Given  C=  80°  47',  Z>  =  2103.2. 

34.  Given   a  =79.24,  c=  106.62. 

35.  Given  A  =  70°  19',  c  =  .5623. 

36.  Given  C'=  151°  28',  c  =  95.47. 

37.  A  regular  pentagon  is  inscribed  in  a  circle  wliose 
diameter  is  24  inches.     Find  the  length  of  its  side. 

38.  At  a  distance  of  100  feet  from  the  base  of  a  tower, 
the  angle  of  elevation  of  its  top  is  observed  to  be  38°.  Find 
its  height. 

39.  What  is  the  angle  of  elevation  of  the  sun  when  a 
tower  wliose  height  is  103.7  feet  casts  a  shadow  1G7.3  feet 
in  length? 

40.  If  tlie  diameter  of  a  circle  is  3268,  find  the  angle  at 
the  centre  subtended  by  an  arc  whose  chord  is  1027. 

41.  If  the  diameter  of  the  earth  is  7912  miles,  what  is  the 
distance  of  the  remotest  point  of  the  surface  visible  from  the 
summit  of  a  mountain  1^  miles  in  height? 

42.  Find  the  length  of  the  diagonal  of  a  regular  pentagon 
whose  side  is  7.028  inches. 

43.  What  is  the  angle  of  elevation  of  a  mountain-slope 
which  rises  238  feet  in  a  horizontal  distance  of  one-eighth  of 
a  mile? 

44.  From  the  top  of  a  lighthouse,  133  feet  above  the  sea, 
the  angle  of  depression  of  a  buo}^  is  observed  to  be  18°  25'. 
Required  the  horizontal  distance  of  the  buoy. 


72  PLANE   TRIGONOMETRY. 

45.  A  ship  is  sailing  due  east  at  the  rate  of  7.8  miles  an 
hour.  A  headland  is  observed  to  bear  due  north  at  10.37 
A.M.,  and  33°  west  of  north  at  12.43  p.m.  Find  the  distance 
of  the  headland  from  each  point  of  observation. 

46.  If  a  chord  of  41.36  feet  subtends  an  arc  of  145°  37', 
what  is  the  radius  of  the  circle  ? 

47.  The  length  of  the  side  of  a  regular  octagon  is  12 
inches.  Find  the  radii  of  the  inscribed  and  circumscribed 
circles. 

48.  How  far  from  the  foot  of  a  pole  80  feet  high  must  an 
observer  stand,  so  that  the  angle  of  elevation  of  the  toj)  of 
the  pole  may  be  10°? 

49.  If  the  diagonal  of  a  regular  pentagon  is  32.83  inches, 
what  is  the  radius  of  the  circumscribed  circle  ? 

50.  From  the  top  of  a  tower,  the  angle  of  depression  of 
the  extremity  of  a  horizontal  base  line,  1000  feet  in  length 
measured  from  the  foot  of  the  tower,  is  observed  to  be 
21°  16' 37".     Find  the  height  of  the  tower. 


o 


51.  If  the  radius  of  a  circle  is  723.29,  what  is  the  length 
of  the  chord  which  subtends  an  arc  of  35°  13'? 

52.  A  regular  hexagon  is  circumscribed  about  a  circle 
whose  diameter  is  10  inches.     Find  the  length  of  its  side. 

53.  From  the  top  of  a  lighthouse,  200  feet  above  the  sea, 
the  angles  of  depression  of  two  boats  in  line  with  the  light- 
house are  observed  to  be  14°  and  32°  respectively.  What  is 
the  distance  between  the  boats? 

54.  A  ship  is  sailing  due  east  at  a  uniform  rate  of  speed. 
At  7  A.M.,  a  lighthouse  is  observed  bearing  due  north,  10.32 
miles  distant,  and  at  7.30  a.m.  it  bears  18°  13'  west  of  north. 
Find  the  rate  of  sailing  of  the  ship  and  the  bearing  of  the 
lifi-hthouse  at  10  a.m. 


SOLUTION  OF   RIGHT   TRlAXGLES. 


FORMULA   FOR  THE  AREA   OF   A  RIGHT   TRIANGLE. 
111.   Case  I.    Given  the  hypotenuse  and  an  acute  angle. 

B 


Denoting  the  area  by  A",  we  have  by  Geometry, 

2K=ab. 
But  by  Art.  L3,     a  =  c  sin  A, 
and  6  =  ceos^. 

Whence,  27r=c^sinYl  cos^l  =  ^c^8in2^  (Art.  74). 

That  is,  4  K=z  r  sin  2  A.  (36) 

In  like  manner, 

4  7r=c-sin2^.  (37) 

Case  II.    Given  an  angle  and  its  opposite  side. 

We  have,      cot  ^1  =  -•»    or     h  =  a  cot  A. 

a 

Whence,  2  K=  a  •  a  cot  A  =  a-  cot^.  (38) 

In  like  manner, 

2  K=h- cot  B.  (39) 

Case  III.    Given  an  angle  and  its  adjacent  side. 
We  have,      cot  A  =  tan  B  (Art.  14) . 
Whence  by  (38) , 

2  7r=a-tan^.  (40) 

In  like  manner, 

27r=?>-tan.4.  (4l) 


74  PLANE  TRIGONOMETRY. 

Case  IV.    Given  the  hypotenuse  and  another  side. 

Since  a^  -}-  6^  =  c^,  we  have  _:i- 

=  a  v/(c  +  a  )  (c  -  a).  (42) 

In  like  manner, 

2/ir=5\/(c  +  6)(c-6).  (43) 

Case  V.    Given  the  tivo  sides  about  the  right  angle. 

In  this  case,     2K=ab.  (44) 

EXAMPLES. 
112.    1.    Given  c  =  10.36,  B=lo° ;  find  the  area. 
By  (37),  4K=:rsm2B. 

Whence,  log  (4^)  =  2  logc  +  log  sin  25. 

lege  =  1.0153  ;  multiply  by  2  =  2.0306 
2B=  150° ;  log  sin  =  9.6990  -  10 

log(4^)=  1.7296 

.-.  4.K=  53.65,  and  K=  13.41. 

Note.     To  find  log  sin  150^,  take  either  log  cos  60=  or  log  sin  30°. 
(See  page  10  of  the  explanation  of  the  tables.) 

Find  the  areas  of  the  followinsj  trianoles  : 

2.  Given  ^=19°  36',  a  =  22.17. 

3.  Given  B  =  24°  7',  a  =  .8213. 

4.  Given  a  =149.31,  5  =  76.29. 

5.  Given   6  =.3056,  c=.6601. 

6.  Given  ^  =  30°  56' 20",  c=  192.9. 

7.  Given  A  =  58°  52',  b  =  .05207. 

8.  Given   a  =.932,  c=  2.786. 

9.  Given  B  =  72°  25',  c  =  27.283. 
10.  Given  5=29°  18'  15",  6=  .33784. 


GENEKAL  PROPERTIES  OF  TRIANGLES. 


75 


VIII.    GENERAL    PROPERTIES   OP 
TRIANGLES. 

113.    In  any  triangle^  the  sides  are  proportional  to  the  sines 
of  their  opposite  angles. 


There  will  ))e  two  cases,  according  as  the  angles  are  all 
acute  (Fig.  1),  or  one  of  them  obtuse  (Fig.  2). 
In  each  case  let  CD  be  drawn  perpendicular  to  AB, 
Then  in  either  figure,  we  have 


sin  A  = 


CD 


In  Fig.  1,  sin  J5  = > 

a 

and  in  Fig.  2,  sin  B  =  sin  (180°-  CBD) 

=  sin  CBD  (Art.  47) 

^CD 

a 

Dividing  these  equations,  we  have  in  either  case 

CD 

sin  A  _  _b__  _  a 
sin  B'~'CD~b 
a 


(45) 


76  PLANE   TRIGONOMETRY. 

In  like  manner,  we  may  prove 

sin  B  _h 
sin  C      c 


(*6) 


,  Sin  C      c  /._N 

and  - — 7=—  (47) 

sm  A      a 

The  above  results  may  be  expressed  more  compactly  as 


sm 


A      sin  B      sin  C 


114.  hi  any  triangle^  the  sum  of  any  two  sides  is  to  their 
difference  as  the  tangent  of  half  the  sum  of  the  opposite  angles 
is  to  the  tangent  of  half  their  difference. 

Formula  (45)  of  Art.  113  may  be  put  in  the  form 

a:b  =  sin  A  :  sin  B. 

Whence,  by  composition  and  division, 

a  +  6  :  a  —  &  =  sin  ^  4-  sin  5 :  sin  ^  —  sin  B, 

a  +  5  _  sin  A  +  sin  B 
'  a  —  6      sin  ^  —  sin  5 

But  by  Art.  73, 

sin .^  +  sin  J5 _  tan |(^  +  B) 
sin  A  —  siuB     tan  ^{A  —  B) 

„„  a-\-b      tan-i-(^4--S)  /-_x 

Whence,  -^  =  - — I)  .      -,,('  (48) 

a  —  b      tani(^  — ij) 

In  like  manner,  we  may  prove 


or 


6  +  c^  tan  jrjB-j-C) 
6-c~tani(J5-(7) 


(49) 


and  c  +  a.^tani(0  +  ^).  (^^^ 

c  —  a      tan^(C  — -4) 


GENERAL  PROPERTIES  OF  TRIANGLES. 


77 


115.    Since  A  +  B=\SO°-  C,  we  have 

tani(^  +  ^)  =tan(90°-iC)  =  coti(7  (Art.  14). 
Thus  formula  (48)  may  be  put  in  the  form 
a-{-b  cot  ^  C 


a  —  b      tan  ^  (^4  —  B) 


(51) 


116.  /n  cmy  triangle,  the  square  of  any  side  is  equal  to  the 
sum  of  the  squares  of  the  other  two  sides,  minus  twice  their 
product  into  the  cosine  of  their  included  angle. 

Case  I.    When  the  included  angle  A  is  acute. 


There  will  be  two  cases,  according  as  the  remaining  angles 
are  both  acute  (Fig.  1),  or  one  of  them  obtuse  (Fig.  2). 
In  each  case  let  CD  be  draAvn  perpendicular  to  AB. 

Then  in  Fig.  1,  BD  =  c  -  AD, 


and  in  Fig.  2, 


BD  =  AD-  c. 


Squaring,  we  have  in  either  case, 

BIX  =  jW'  +  c2  _  2c  X  AD. 

2 

Adding  CD'  to  both  members, 

mf  +  ~Clf  =  A&  +CD^-\-(?-2cy.AD. 
But,  BD'  -{-W=  a-,  and  AD'  +  CD^  =  h\ 


Also, 
Therefore, 


cos  A  = 1   or   AD  =  b  cos  A. 

b 

a^  =  b-  -\-c-  —  2bc  cos  A. 


(52) 


78  PLANE   TlilGONOMETRY. 

Case  II.    When  the  included  angle  A  is  obtuse. 


In  Fig.  3,         BD  =  AD  +  c. 

Squaring,  and  adding  OD'  to  both  members, 

Blf+Clf  =  Aff  4-  OS'  +  c^  +  2  c  X  AD. 


But,     BD^  +  CD"  =  a\  and  AD"  +  CD"  =  Ir. 
Also,  cos  A  =  cos  (180°  -  CAD) 

=  -  cos  CAD  (Art.  47) 
AD 


Whence,  AD  —  —  b  cos  A. 

Therefore,  a^  =b'--^c^-2bc  cos  A. 

In  like  manner,  we  may  prove 

b^  =  c^  -\-  a^  —  2  ca  cos  5, 


and 


c^  =  a^  -|-  &2  _  2  a6  cos  C 


(53) 
(54) 


117.    To  express  the  cosines  of  the  angles  of  a  triangle  in 
terms  of  the  sides  of  the  triangle. 


From (52),  Art.  116, 


a 


2_  7.2 


Whence, 


6^  +  c^  —  2  6c  cos  A. 


or, 


2  be  cos  J.  =  6-  +  c^  —  a^, 

[>2  ^  ^.-  _  ^^2 


cos  A 


2  be 


(55) 


GEXERAL   PROrERtlES  OF   TRIANGLES.  79 

Id  like  manner,  we  have 

cosB  =  — '— ,  (56) 

and  cos  C  =  — (57) 

118.    To  express  the  sines,  cosines,  and  tangents   of  the 
ha/f-angles  of  a  triangle  in  terms  of  the  sides  of  the  triangle. 

From  (55),  Art.  117, 

cos  A  =  — 

26c 

Subtracting  both  members  from  unity, 

.  <       1       ^"  +  c"  —  a-      a-—  6- -f  2bc  —  c^ 

1  —  cos  ^1=1 ■ = ■ 

2  be  2  be 

Whence  by  (a),  Art.  75, 

2  sm-  ^A  = r-^ 5 

^  2  be 

(a  —  b-\-c)(a-\-b  —  c) 
or,  sin^  IA  =  ^ f~^ -- 

^  Abe 

Denoting  «  +  6  +  c  by  2  s,  so  that  s  is  the  half-sum  of  the 
sides  of  the  triangle,  we  have 

a  -  b  -\-  c  =  (a-^b  -he)  -  2b  =  2 s-2h  =  2  (s-b) , 

and  a  4-  6  —  c  =  {a-{-b  -f-c)  —  2 c=  2  s—2 c  =  2  {s  —  c) . 

4(s-6)  (s-c) 


Hence,        sin^4^  = 


4  be 


.    ■,    J         lis  —  b)(s  —  c)  .     . 

or,                    sin|-^=^^^ ^-J^ ^.  (58) 

In  like  manner,  we  may  prove 

sm^B-\^ — ,  (59) 


and 


•     1  ^        i(s  —  a)  (s  —  b  .      ^ 

smiC=sj^ ^ (60) 


or,  cos^^A  = 


80  PLANE   TRIGONOMETRY. 

Again,  adding  both  members  of  (55)  to  unity,  we  have 

l  +  eos^  =  l  +  -l^^^  = Uc 

Whence  by  (a),  Art.  75, 

(6  +  c  +  a)  (6  +  c  — ct). 
45c 
But,         64-c  +  a=2s, 

and  b-{-c  —  a  =  {b-\-c  +  a)  —  2a  =  2(s  —  a), 

„ ,    .      4  s  (s  —  a) 
Hence,       cos^t^  = tt ' 

C0si4=JS.  (61) 

In  like  manner, 

cosii3=Ji5HS,  (62) 

"^  \       ca 


and  cosiO=J^^^-  (63) 

Dividing  (58)  by  (ei) ,  we  obtain 


be 


s{s  —  a) 


= >-;)(^-^).  (64) 

\       s{s  —  a) 
In  like  manner, 

n     -r.  !(S—    C)     (S  —  a)  /_^N 

fa>"i-s=^^  .(;_ft)  '  («=) 


,  ^        l(s  — a')(s  — 6)  /_->. 

and  t^niC^^^^^^^.  (66) 

Note.  Since  each  angle  of  a  triangle  is  less  than  180'',  its  half  is 
less  than  90^ ;  hence  the  positive  sign  must  be  taken  before  the  radical 
in  each  of  the  formulae  of  Art.  118. 


GENERAL   PROPI^RTIES  OP   TRIANGLES. 


81 


AREA   OF   AN  OBLIQUE  TRIANGLE. 
119.     Case  I.   Given  two  sides  and  their  included  angle. 

C 


Fig.  1. 

There  will  be  two  cases,  according  as  the  included  angle 

A  is  acute  (Fig.  1),  or  oljtuse  (Fig.  2). 

In  each  case  let  CD  ijc  drawn  perpendieuhir  to  ^17^. 

Then,  denoting  the  area  of  the  triangle  by  A",  we  have  by 

Geometry, 

2K=rx  CD. 

But  in  Fio-.  1 ,  sin  A  = ? 

b 

and  in  Fig.  2,       sin  A  =  sin  (180° -CAD) 


=  sin  CzlZ)  (Art.  47)  = 


CD 


Whence,  in  either  figure, 

CZ>  =  6sin^4. 
Therefore ,  2  A"  =  be  sin  A. 

In  like  manner, 

2Il  =  ca  sin  B, 
and  2  K=  ab  sin  C. 

Case  II.     Given  a  side  and  all  the  angles. 
By  (69) ,  2  K=  ab  sin  C. 

But  by  Art.  118, 

b      sin  B  y      a  sin  5 

or   b  == 


(67) 

(68) 
(69) 


a      sin  ^1 


sin  A 


:^Lane  trigonometry. 


Substituting,    ^K—  a  x X  sin  G 

sin  A 

o?  sin  B  sin  C 


sin^ 


In  like  manner, 


(70) 


o  -r^     IP'  sin  C  sin  ^  ,      , 

2/1  = —,  (71^ 

sm^ 

1                         c\  rp-     ^  sin  A  sin  5  ,     . 

and  1K= ^—- (72) 

sni  G 

Case  III.     Given  the  three  sides. 

By  (67),  2/r=6csin^ 

=  2  be  sin  ^A  cos  ^A  (Art.  74) . 

Dividing  by  2,  and  substituting  the  values  of  sin|^^  and 
cos^^  from  Art.  Il5,  we  have 


h'^=hr,.j(s-b)(s-c)     Is(s-a) 
\  be  \       be 


=  V/s(s  — a)(s  — 6)(s  — c).  (73) 


SOLUTIOI^  OF   OBLIQUE   TRIANGLES. 


IX.    SOLUTION    OF    OBLIQUE   TRIANGLES. 

120.  Ill  the  solution  of  plane  oblique  triangles  we  may 
distinguish  four  cases : 

1 .  Given  a  side  and  any  tivo  angles. 

2.  Given  two  sides  and  their  included  angle. 

3.  Given  the  three  sides. 

4.  Given  t/co  sides  and  the  angle  opposite  to  one  of  them. 

Case  I. 

121.  Given  a  side  and  any  two  angles. 

The  third  angle  may  be  found  by  Geometry,  and  then  by 
aid  of  Art.  113  the  remaining  sides  may  be  calculated. 

Tlie  triangle  is  always  possible  for  any  values  of  the  given 
elements,  provided  the  sum  of  the  given  angles  is  <  180°. 

1.    Given  b  =  20,  .1  =  104°,  B  =  19°.     Find  «,  c,  and  C. 

C=  180°  -  (.1  -f-  B)  =  l.SO°  -  123°  =  57°. 

■D     A  4.    1 1  o    <^<'      sin  A             T          c      sin  C 
By  Art.  113,  -  = ,        and  -  = 

b       sinB  b      sin  J5 

That  is,  a  =  b  sin  A  esc  B,  and  c  =  b  sin  C  esc  B. 

Whence,    log  a  =  log  b  -f  log  sin  A  -\-  log  esc  B, 
and  log  c  =  log  b  -j-  log  sin  C  +  log  esc  B. 

log  6  =1.3010  \ogb^  1.3010 

log  sin  ^1=  9.9869  -  10  log  sin  C=  9.9236  -  10 

log  CSC  B  =  0.4874  log  esc  B  =  0.4874 

loora  =  1.7753  loo:  c=  1.7120 

.-.  a  =  59.61.  .-.   c  =  51.52. 

Note.  To  find  the  log  cosecant  of  an  angle,  subtract  the  log  sine 
from  10  —  10.  To  find  log  sin  104°,  take  either  log  cos  14°  or  log  sin 
76°.     (See  pages  7  and  10  of  the  explanation  of  the  tables.) 


84  PLANE   TRIGONOMETRY. 

EXAMPLES. 
Solve  the  following  triangles  : 

2.  Given  a  =10,  ^  =  38°,  5  =77°  10'. 

3.  Given  &  =  .8037,     5  =  52°  20',  (7  =101°  40'. 

4.  Given  c  =  .032,       ^  =  36°  8',  5  =  44°  27'. 

5.  Given  6  =  29.01,     ^=87°  40',  (7  =  33°15'. 

6.  Given  ct  =  5.42,       5  =  98°  22',  C  =  41°l'. 

7.  Given  c  =  .0161,     ^  =  35°  15',  (7=  123°  39'. 

8.  Given  a  =  400,        ^  =  54°  28',  0=60°. 

9.  Given  6  =  314.29,  .4  =  67°  22',  5  =  57°  51'. 
10.    Given  c  =  7.86,       5  =  32°  2' 52",  0  =  43°  25' 26". 

Case  II. 

122.    Given  two  sides  and  their  included  angle. 

Since  one  angle  is  known,  the  sum  of  the  remaining  angles 
may  be  found,  and  then  their  difference  may  be  calculated 
b}^  aid  of  Art.  114. 

Knowing  the  sum  and  difference  of  the  angles,  the  angles 
themselves  may  be  obtained,  and  then  the  remaining  side 
may  be  computed  as  in  Case  I. 

The  triangle  is  possible  for  any  values  of  the  data. 

1.    Given  a  =  167,  c  =  82,  5  =  98°.     Find  A,  O,  and  6. 

By  Geometry,  ^  +  O  =  180°  -  5  =  82°. 

By  Art.  114,  o  +  c^  tanJC^  +  0)^ 

a  —  c      tan^(^— O) 

or,  tanl-(^-0)=-^^  tan  1(^  +  0). 

Whence,     log  tan  |(^  —  O)  =  log  {a  —  c) -\-  colog  (a  -f-  c) 

+  logtani(^  +  0). 


SOLUTION   OF   OBLIQUE   TRIANGLES.  85 

a-c  =  85  log  =1.9294 

a  +  c  =  249  colog  =  7.6038 

i(A  +  C)  =  41°  log  tan  =  9.9392 

logtan^(^-C)=  9.4724 
.-.  i{A-C)=l6°Sl.T. 
Hence,  A  =  ^{A  +  C)  +  \{A  -C)  =  57°  31.7', 

and  C  =  i{A  -\- C)  -  ^{A- C)=  24°  28.3'. 

To  find  the  remaining  side,  we  have  by  Art  113, 

,       a  sin  B  ■     n         a 

h  z=z =  a  sin  B  CSC  A. 

sin  A 

Whence,  log  h  =  log  a  +  log  sin  B  +  log  esc  A. 

loga  =  2.2227 
log  sin  2^  =9.9958 
log  CSC  ^1  =  0.0739 

log  6  =  2.2924 
.-.  6=196.05. 

EXAMPLES. 
Solve  the  following  triangles  : 

2.  Given  c/  =  27,  c=15,  J5  =  46°. 

3.  Given  a  =  486,        6  =  347,         0=51°  36'. 

4.  Given  6  =  2.302,     c  =  3.567,     ^=62°. 

5.  Given  a  =.3,  6  =  .363,        0=  124°  56'. 

6.  Given  5=1192.1,   c  =  356.3,     ^  =  26°  16'. 

7.  Givena=7.4,         c  =  11.439,  5  =  82°  26'. 

8.  Given  a  =  53.27,     6  =  41.61,      (7=  78°  33'. 

9.  Given  6  =  .02668,   c=. 05092,  ^=115°  47'. 
10.    Given  a  =  51.38,*  c=  67.94,     ^=79°  12' 34". 


86  PLANE  TKIGONOMETRY. 

Case  III. 
123.    Given  the  three  sides. 

The  angles  might  be  calculated  by  the  formulae  of  Art. 
117;  but  as  these  are  not  adapted  to  logarithmic  computa- 
tion, it  is  more  convenient  to  use  the  formulae  of  Art.  118. 

Each  of  the  three  angles  should  be  computed  trigonometri- 
call}^  as  we  then  have  a  check  on  the  work,  since  their  sum 
should  be  180°. 

If  all  the  angles  are  to  be  computed,  the  tangent  formulae 
are  the  most  convenient,  as  only  four  different  logarithms 
are  required.  If  but  one  angle  is  required,  the  cosine  for- 
mulae will  be  found  to  involve  the  least  work. 

The  triangle  is  possible  for  any  values  of  the  data,  pro- 
vided no  side  is  greater  than  the  sum  of  the  other  two. 

If  all  the  angles  are  required,  and  the  tangent  formulae  are 
used,  they  may  be  conveniently  modified  as  follows : 


^     .           ^            -.J         \(s  —  a)(s  —  b)(  s  —c) 
By  Art.  118,    tani^=^^^ s\s-ay 

l_    Ki^a){s-b){s-c) 

s  —  a\  s 


^         .  \(s  —  a)(s —  b)  (s  —  c)  , 

Denotmg  lI- — — by  r,  we  have 


tan^^  = -. 

s  —  a 


In  like  manner, 

tan  ^B  = ,  and  tan  |-  (7  = 

s—b  s—c 

1.    Given  a  =  2.5,  6  =  2.79,  c  =  2.33  ;  find  A,  B,  and  G. 

In  this  case,  2s  =  a  -f-  5  +  c  =  7.62,  and  s=  3.81. 

Whence,  s  —  a=  1.31,  s  —  b  =  1.02,  and  s  —  c=  1.48. 

By  logarithms,  we  have 
log  r  =  -2-[log  (s  —  a)-]-  log  (s  —  6)  +  log  (s  —  c)  +  colog  s] . 


SOLUTION   OF   OBLIQUE   TRIANGLES.  87 

Also,  log  tan  ^A  =  log  r  —  log  (s  —  a) , 
log  tan  ^B  =  log  r  —  log  (s  —  6) , 
log  tan  ^C  =  log  r  —  log  (s  —  c) . 

log  (s  -  a)  =  0.11 73  log  r  =  9.8576  -  10 

log  (s  -  h)  =  0.0086  log  (s  -  6)  =  0.0086 


log  (s  -  c)  =  0.1703  log  tan  4^i?=  9.8490  -  10 
cologs  =  9.4191  -10  ^5=35°  14.1' 

2)9.7153-10  .'.B=  70°  28.2'. 

.•.logr=  9.8576 -10 

log  r  =  9.8576  -  10  log  r  =  9.8576  -  10 

log  (s  -  a)  =  0.1173  log  (s  -  c)  =  0.1703 

log  tan  ^A  =  9.7403  -  10  log  tan  ^C=  9.6873  ^  10 

iA  =  28°  48.3'  i C=  25°  57.2' 

.-.  A  =  57°  36.6'.  .-.•0=  51°  54.4'. 

Check,  ^14-J5  +  C=179°59.2'. 

2.    Given  a  =  7,  6  =  11 ,  c  =  9.6  ;  find  B. 


Ms-b) 


By  Art.  118,   cos^^ 

Or,    log  cos^B  =  4-[log  s  +  log  {s  —  b)-\-  colog  c  +  colog  a], 
In  this  case,  2s  =  a -|- 6  +  c  =  27.6. 

Whence,  s=13.8,  and  s-b  =  2.8. 

logs  =1.1399 
log  (s  -  6)  =  0.4472 
colog  c  =  9.0177 -10 
colog  a  =  9. 1549  — 10 
2)19.7597-20 
log  cos  1^=9.8798 -10 
1^=40°  41.8' 
r.B  =  8r  23.6'. 


88  PLANE   TRIGONOMETRY. 

EXAMPLES. 
Solve  the  following  triangles  : 

3.  Given  a  =  2,  &  =  3,  c  =  4. 

4.  Given  a  =  4,  5=7,  c=6. 

5.  Given  a  =  5. 6,  5  =  4.3,  c  =  4.9. 

6.  Given  a=. 23,  5=.26,  c=.198. 

7.  Given  a  =79.3,  6=94.2,  c  =  66.9. 

8.  Given  a  =  321,  6  =  361,  c  =  402. 

9.  Given  «=. 641,  6=.529,  c=.702. 
10.  Given  a  =  3.019,  6  =  6.731,  c  =  4.228. 

Case  IY. 
124.    Given  two  sides  and  the  angle  opposite  to  one  of  them. 

It  was  stated  in  Art.  106  that  a  triangle  is  in  general  com- 
pletely determined  when  three  of  its  elements  are  known, 
provided  one  of  them  is  a  side.  The  only  exceptions  occur 
in  Case  IV. 

To  illustrate,  let  us  consider  the  following : 

1.  Given  a  =  52.1,  6  =  61.2,  ^  =  31°  26'.  Required  B, 
(7,  and  c. 

-n     A  ^   -.  1  o                sin  B     i>             .    ^      6  sin  ^ 
By  Art.  113,  - — -  —  —>    or   sinB  = 

Whence,  log  sin  B  =  log  6  +  colog  a  +  log  sin  A. 

log  6  =1.7868 
cologa  =  8.2832 -10 
log  sin  J.  =  9.7173 -10 


log  sin  5  =  9.7873 -10 

.-.  ^  =  37°  47.5',  from  the  table. 


SOLUTION   OF   OBLIQUE   TRIAXGLES. 


89 


But  in  determining  the  angle  corresponding,  attention 
must  be  paid  to  the  fact  that  an  angle  and  its  supplement 
have  the  same  sine  (Art.  47). 

Therefore  another  value  of  5  tvIII  be  180°  —  37°  47.5',  or 
142°  12.5' ;  and  calling  these  values  Bi  and  ^o.  we  have 

^1=37°  47.5',  and  5,=  142°  12.5'. 

Note.  The  reason  for  tliis  ani})iguity  is  at  once  apparent  when  we 
attempt  to  construct  the  triangle  from  the  data. 


Bi  D 


We  first  lay  off  the  angle  DAF  equal  to  31°  2(V,  and  on  AF  take 
^C— 61.2.  With  C  as  a  centre,  and  a  radius  equal  to  52.1,  describe 
an  arc  cutting  AD  at  Bi  and  B.^.  Then  either  of  the  triangles  AB^C 
or  vlC^C  satisfies  the  given  conditions. 

Tlie  two  values  of  B  which  were  obtained  are  the  values  of  the 
angles  AB^C  and  AB.,C  respectively;  and  it  is  evident  geometrically 
that  these  angles  are  supplementary. 

To  complete  the  solution,  denote  the  angles  ACB^  and 
ACBo  by  C\  and  02,  and  the  sides  ABi  and  xlB.,  b}-  q  and  Cg. 

Then,  Ci=  180°- (.1  +  7?,)  =  180°-  69°  13.5'=  110°  46.5', 
and  O2=180°-(.l  +  i?,,)  =  180°-173°38.5'=     6°  21.5'. 


.      .  c,      sin  C, 

Agam,      -L  =  -^-  ^ 

a      sm  A 


and 


"Whence,       c,  =  a  sin  d  csc^, 
log  a  =1.7168 
log  sin  ^1  =  9.9708 
log  csc^  =  0.2827 


^2. 

a 

C2 


sin  Co 


sin  .4 
and        C2  =  a  sin  C^  esc  A. 

log  a  =1.7168 
log  sin  a  =  9.0443 
log  csc^  =  0.2827 


logCi=  1.9703 
.-.  Ci=93.4. 


log  Co  =1.0438 
.-.  C2=  11,06. 


90  PLANE   TEIGOJs^OMETRY. 

125.  AYhenever  an  angle  of  an  oblique  triangle  is  deter- 
mined fr.om  its  sine,  both  tiie  acute  and  obtuse  values  must 
be  retained  as  solutions,  unless  one  of  them  can  be  shown  by 
other  considerations  to  be  inadmissible ;  and  hence  there 
may  sometimes  be  two  solutions,  sometimes  only  one,  and 
sometimes  none,  in  an  example  under  Case  IV. 

I.    Let  the  data  be  a,  6,  and  A,  and  suppose  b<.ct- 
Since,  by  Geometry,  B  must  be  <  A,  only  the  acute  value 
of  B  can  be  taken ;  in  this  case  there  is  but  one  solution. 

II.    Let  the  data  be  a,  b,  and  A,  and  suppose  b  >  a. 

Since  B  must  be  >  A,  the  triangle  is  impossible  unless  A 
is  acute. 

.      .        .         sin  B      b         17-^  -Tj-^-.* 

Agam,  smce =  -5  and  0  is  >  a,  smB  is  >  sin  J.. 

sin  A      a 

Hence  both  the  acute  and  obtuse  values  of  B  are  >  A, 
and  there  are  tico  solutions,  except  in  the  following  cases  : 

If  the  data  are  such  as  to  make  log  sin -5  =  0,  then 
sin 5=  1  (Art.  87)  and  B=  90°,  and  the  triangle  is  a  right 
triangle  ;  if  log  sin  5  is  positive,  then  sin  5  is  >  1,  and  the 
triangle  is  impossible. 

126.  The  results  of  Art.  125  may  be  stated  as  follows : 

If,  of  the  given  sides,  that  adjacent  to  the  given  angle  is 
the  less,  there  is  but  one  solution,  corresponding  to  the  acute 
value  of  the  opposite  angle. 

If  the  side  adjacent  to  the  given  angle  is  the  greater,  there 
are  two  solutions  unless  the  log  sine  of  the  opposite  angle  is 
0  or  positive  ;  in  which  cases  there  are  one  solution  (a  right 
triangle),  and  no  solution,  respectively. 

127.  We  will  illustrate  the  above  points  by  examples  : 

2.    Given  a  =7.42,  6  =  3.39,  ^=105°  13';  find  5. 
Since  ?>  is  <  o,  there  is  but  one  solution,  corresponding  to 
the  acute  value  of  B. 


SOLUTIOX   OF   OBLIQUE   TRIANGLES.  91 

Air    V.  •     13      ^sin^ 

We  nave,        sin  B  = • 

a 

log  6  =  0.5302 

colog  a  =  9.1296 

log  sin  .4  =  9.9845 


log  sin  5  =  9.6443 
.-.  i3=2G°  9.0'. 

3.  Given  &  =  3,  c=2,   C=100°;  find  J5. 

Since  h  is  >  c,  and  C  is  obtnse,  the  triangle  is  impossible. 

4.  Given  rt=  22.764,  c  =  50,  ^1  =  27°  4.8';  find  C. 

We  have,         sin  C  =  — 

logc=  1.G990 

colog  a  =  8.G428 

log  sin  .4  =  9.6582 


log  sin  C'=  0.0000 
.-.  sin  C=  1,  and  C=90°. 
Here  there  is  bnt  one  solntion  ;  a  right  triangle. 

5.    Given  a  =  .83,  6  =.715,  7^=61°  47';  find  ^. 

WT    \^                       A      ft  sin  B 
We  have,        sin  A  = 

h 
log  rt  =  9.9191 

colog  &  =  0.1457 

W  sin  5=  9.9451 


log  sin  .1  =  0.0099 
Since  log  sin  A  is  positive,  the  triangle  is  impossible. 

EXAMPLES. 
Solve  the  followinor  triansiles  : 

6.  Given  a  =  5.08,  5  =  3.59,  ^=63°  50'. 

7.  Given  5  =  74.8,  c  =  02.2,  0=27°  18'. 


92  PLANE   TRIGONOMETRY. 

8.  Given  6  =  .2337,         c  =  .1982,         5  =  109°. 

9.  Given  a  =1.07,  c  =  1.71,  0  =  31°  53'. 

10.  Given  a  =.1864,  5  =  .17,  5  =  63°  40'. 

11.  Given  a  =  50,  c  =  66,  ^=123°11'. 

12.  Given  6  =  50.3,  c  =  66.8,  0  =  32°  49'. 

13.  Given  a  =  8.656,  c  =  10,  ^=59°  57'. 

14.  Given  6  =  5.161,  c  =  6.84,  J5=44°3'. 

15.  Given  a  =  214.56,  6  =  284.79,  5  =104°  20'. 

16.  Given  6  =  3069,  c  =  1223,  0  =  55°  52'. 

17.  Given  a  =  .7097,  c  =  .5112,  ^  =  35°  11'. 

18.  Given  a  =106. 85,  6  =  166.21,  ^  =  40°0'21". 

19.  Given  a=. 3216,  c  =  .2708,  0=52°24'. 

20.  Given  6  =  811.3,  c  =  606.4,  jB  =  126°  5' 20". 

AREA   OF  AN   OBLIQUE   TRIANGLE. 

128.    1.  Given  a  =18.063,  ^  =  96°  30',  5=35°;  find  A". 

_      .            ^      ^  ^^     a^  sin  B  sin  O 
By  Art.  119,    2A  = -. — -. 

-^  '  sm  ^ 

=  a^  sin  B  sin  0  esc  A. 
Whence, 

log  (2  K)  =  2  log  a  +  log  sin  B  +  log  sin  O  +  log  esc  A. 

From  the  data, 

0  =  180°  -{A  +  B)  =  48°  30'. 

log  a  =  1.2568  ;  multiply  by  2  =  2.5136 

log  sin  5  =  9.7586 
log  sin  0  =  9.8745 
log  CSC  ^=0.0028 

log  (2Ar)  =  2.1495 

.-.2^=141.1,  and  ^=70.55. 


SOLUTION  OF   OBLIQUE   TRIANGLES.  93 

EXAMPLES. 
Find  the  areas  of  the  following  triangles  : 

2.  Given  rt  =  38,  c=G1.2,  5  =  67°  56'. 

3.  Given  a  =5,  5  =  7,  c  =  6. 

4.  Given  6  =  2.07,  ^=70°,  5  =  36°  23'. 

5.  Given  /j  =  116.1,  c  =  100,  ^  =  118°16'. 

6.  Given  a  =79,  b  =  \)-i,  c=67. 

7.  Given  a  =  3.123,         ^  =  53M1',  5=13°57'. 

8.  Given  /j  =  .439,  ^1  =  76°  38',  (7=  40°  35'. 

9.  Given  a  =  23.1,  6  =  19.7,  c=25.2. 

10.  Given  «  =  . 3228,  c  =  .9082,  i^  =  60°lG'. 

11.  Given  r  =80.25,  7?  =100°  5',  0=31°  44'. 

12.  Given  (<  =  .010168,  6  =  .018225,  C  =  ll°18'26". 

13.  Given  (t  =  5.82,  6  =  6,  c=4.26. 

MISCELLANEOUS    EXAMPLES. 

129.  1.  From  a  point  in  the  same  horizontal  plane  with 
the  base  of  a  tower,  the  angle  of  elevation  of  its  top  is 
52°  39',  and  from  a  point  100  feet  fin-ther  away  it  is  35°  16'. 
Required  the  height  of  the  tower,  and  its  distance  from  each 
of  the  points  of  observation. 

2.  In  a  field  ABCD,  the  sides  AB,  BC,  CD,  and  DA  are 
155,  236,  252,  and  105  rods,  respectively,  and  the  length  of 
a  line  from  xi  to  C  is  311  rods.     Find  the  area  of  the  field. 

3.  From  the  top  of  a  bluff,  the  angles  of  depression  of 
two  posts  in  the  plain  below,  in  line  with  the  observer  and 
1000  feet  apart,  are  found  to  be  27°  40'  and  9°  33',  respec- 
tively.    AYhat  is  the  height  of  the  bluff  above  the  plain? 

4.  Two  yachts  start  at  the  same  time  from  'the  same 
point,  and  sail,  one  due  north  at  the  rate  of  10.44  miles  an 


94  PLAXE   TRIGONOMETRY. 

hour,  and  the  other  due  northeast  at  the  rate  of  7.71  miles 
an  hour.     How  far  apart  are  they  at  the  end  of  40  minutes  ? 

5.  A  ship  is  sailing  due  southwest  at  the  rate  of  8  miles 
an  hour.  At  10.30  a.m.,  a  lighthouse  is  observed  to  bear 
30°  west  of  north,  and  at  12.15  p.m.,  it  is  observed  to  bear 
15°  east  of  north.  Find  the  distance  of  the  lighthouse  from 
each  position  of  the  ship. 

6.  Wishing  to  find  the  distance  of  an  inaccessible  object 
A  from  a  position  B,  I  measure  a  line  BC\  208.3  feet  in 
length.  The  angles  ABC  and  ACB  are  measured,  and 
found  to  be  126°  35'  and  31°  48',  respectively.  Required  the 
distance  AB. 

7.  A  flagpole  40  feet  in  height  stands  on  the  top  of  a 
tower.  From  a  position  near  the  base  of  the  tower,  the 
angles  of  elevation  of  the  top  and  bottom  of  the  pole  are 
38°  53'  and  20°  18',  respectively.  Required  the  distance  and 
height  of  the  tower. 

8.  A  surveyor  observes  that  his  position  A  is  exactly  in 
line  with  two  inaccessible  objects  B  and  C.  He  measures  a 
line  AD,  500  feet  in  length,  making  the  angle  BAD  =  60°, 
and  at  D  observes  the  angles  ADB  and  BDC  to  be  40°  and 
60°,  respectively.     Required  the  distance  BC. 

9.  To  find  the  distance  between  two  buoys  A  and  5,  I 
measure  a  base  line  CD  on  the  shore,  150  feet  in  length. 
At  the  point  C  the  angles  ACD  and  BCD  are  measured  and 
found  to  be  95°  and  70°,  respectively ;  and  at  D  the  angles 
BDC  and  ADC  are  found  to  be  83°  and  30°.  What  is  the 
distance  between  the  buoys  ? 

10.  The  sides  of  a  field  ABCD  are  AB=^1,  BC=63, 
and  DA  =  20,  and  the  diagonals  AC  and  BD  are  75  and  42, 
respectively.     Required  the  area  of  the  field. 


Paet   II. 
SPHERICAL  TPJGOXOMETRY. 


3>*iOO- 


X.    GEOMETRICAL    DEFINITIONS    AND 
PRINCIPLES. 

130.  If  a  triedral  angle  is  formed  with  its  vertex  at  the 
centre  of  a  sphere,  it  intercepts  on  the  surface  a  spherical 
triangle. 

131.  The  triangle  is  bounded  by  three  arcs  of  great  circles 
called  its  sides,  which  measure  the  face  angles  of  the  triedral 
angle. 

The  angles  of  the  spherical  triangle  are  the  diedral  angles 
of  the  triedral  angle  ;  and  by  (ieometry,  each  is  measured 
by  the  angle  between  two  straight  lines  drawn,  one  in  each 
face,  and  perpendicular  to  the  edge  at  the  same  point. 

132.  The  sides  of  a  spherical  triangle,  being  arcs,  are 
usually  expressed  in  degrees. 

If  the  length  of  a  side  in  terms  of  some  linear  unit  is 
desired,  it  ma}^  be  obtained  by  finding  the  ratio  of  its  arc  to 
360°,  and  multiplying  the  result  by  the  length  of  the  circum- 
ference of  a  great  circle. 

133.  Spherical  Trigonometry  treats  of  the  trigonometrical 
relations  between  the  elements  of  a  spherical  triangle  ;  or 
what  is  the  same  thing,  between  the  face  and  diedral  angles 
of  the  triedral  angle  which  intercepts  it. 


96  SPHERICAL   TRIGONOMETRY. 

134.  The  face  and  diedral  angles  are  not  altered  in  mag- 
nitude by  varj^ing  the  radius  of  the  sphere,  and  hence  the 
relations  between  the  sides  and  angles  of  a  spherical  triangle 
are  independent  of  the  length  of  the  radius. 

135.  We  shall  limit  ourselves  in  this  work  to  snch  tri- 
angles as  are  considered  in  Geometry,  where  each  angle  is 
less  than  two  right  angles,  and  each  side  less  than  the  semi- 
circumference  of  a  great  circle  ;  that  is,  where  each  element 
is  less  than  180°. 

136.  The  proofs  of  the  following  properties  of  spherical 
triangles  may  be  found  in  any  treatise  on  Solid  Geometrj' : 

(a)  P^ither  side  of  a  spherical  triangle  is  less  than  the  sum 
of  the  other  two  sides. 

(5)  If  two  sides  of  a  spherical  triangle  are  unequal,  the 
angles  opposite  them  are  unequal,  and  the  greater  angle  lies 
opposite  the  greater  side  ;  and  conversely. 

(c)  The  sum  of  the  sides  of  a  spherical  triangle  is  less 
than  360°. 

(d)  The  sum  of  the  angles  of  a  spherical  triangle  is 
greater  than  1<S0°,  and  less  than  540°. 

(e)  If  A'B'C  is  the  polar  triangle  of  ABC,  i.e.,  if  A, 
B,  and  C  are  the  poles  of  the  arcs  a',  5',  and  c',  respectively, 
then  conversely,  ABC  is  the  polar  triangle  of  A'B'C 


(/)  In  two  polar  triangles,  each  angle  of  one  is  measured 
by  the  supplement  of  the  side  lying  opposite  to  it  in  the 
other. 


GEOMETRICAL   PRINCIPLES. 


97 


That  is, 
A  =  180^ 
A'  =  180^ 


a, 


B  =180° -6', 
B'=  180° -5, 


C  =180°-c', 
(7'  =  180°-c. 


137.  A  spherical  triangle  is  called  tri-rectangular  when  it 
has  three  right  angles  ;  each  of  its  sides  is  a  quadrant,  and 
each  vertex  is  the  pole  of  the  opposite  side. 

138.  I.  Let  C  be  the  right  angle  of  the  spherical  right 
triangle  ABC,  and  suppose  a  <  90°  and  b  <  90°. 

Complete  the  tri-rectangular  triangle  A'B'C. 
Also,  since  B'  is  the  pole  of  AC,  and  A'  of  BC,  construct 
the  tri-rectangular  triangles  AB'D  and  ABE. 

n' 


Then  since  A  and  B  lie  on  the  same  side  of  B'D,  AB  or 
c  is  <  90°. 

Since  BC  is  <  B'C,  the  angle  ^lls  <  B'AD,  or  <  90°. 
Since  AC  is  <  A'C,  the  angle  J5  is  <  A'BE,  or  <  90°. 


II.    Suppose  a  <  90°  and  h  >  90°. 

c  B 


Complete  the  lune  ABA'C. 


98  SPHERICAL   TRIGOXOMETRY. 

Then  in  the  right  triangle  A'BC,  A'C=  180"  -  b. 

That  is,  the  sides  a  and  A'C  of  the  triangle  A'BC  are 
each  <  90°;  and  by  I.,  A'B  and  the  angles  A'  and  A'BC 
are  each  <  90°. 

But,  c  =  180°  -  A'B,  A  =  A',  and  B  =  180°  -  A'BC.     ^ 

"Whence,  c  is  >  90°,  A  <  90°,  and  B  >  90°. 

In  like  maimer,  if  a  is  >  90°  and  &  <  90°,  then  c  is  >  90°, 
^>90°,  and5<  90°, 

III.    Suppose  a  >  90°  and  b  >  90°. 


Complete  the  lune  ACBC. 

Then  in  the  right  triangle  ABC,  AC  =180'' -b,  and 
BC  =  180° -a. 

That  is,  the  sides  AC  and  BC  of  the  triangle  ABC  are 
each  <  90°  ;  and  by  I.,  AB  and  the  angles  BAC  and  ABC 
are  each  <  90°. 

But,  A  =  180°  -  BAC,  and  B  =  180°  -  ABC. 
Whence,  c  is  <  90°,  A  >  90°,  and  B  >  90°. 
Hence,  in  any  spherical  right  triangle : 

1.  If  the  sides  including  the  right  angle  are  in  the  same 
quadrant,  the  hypotenuse  is  <  90° ;  if  they  are  in  different 
quadrants,  the  hypotenuse  is  >  90°. 

2.  An  angle  is  in  the  same  quadrant  as  its  opp)Osite  side. 


SPHERICAL   RIGHT   TRIANGLES. 


99 


XI.    SPHERICAL  RIGHT  TRIANGLES. 

139.    Let  C  be  the  right  anole  of  the  spherical  right  tri- 
angle ABC\  and  0  the  centre  of  the  sphere. 


Join  OA,  OB,  and  OC. 

At  any  })oint  A'  of  OA  draw  A'B'  and  A'C  perpendicular 
to  OA,  and  join  B'C 

Then  hj'  Art.  L'31,  the  sides  «,  6,  and  c  measure  the  angles 
BOC\  COA,  and  AOB,  respectively,  and  the  angle  B'A'C 
is  equal  to  the  angle  A  of  the  spherical  triangle. 

Since  OA  is  perpendicular  to  A'B'  and  A'C,  it  is  perpen- 
dicular to  the  plane  A' B'C. 

AVhence,  since  each  of  tlie  })lanes  A'B'C  and  OBC  is 
perpendicular  to  the  plane  OAC\  their  intersection  B'C  is 
perpendicular  to  OAC. 

Therefore  B'C  is  perpendicular  to  A'C  and  OC. 


In  the  right  triangle  OA'B',  we  have 


cos  c  =  cos  A' OB'  = 


OA' 

OB' 


OC      OA' 
OB'  ^  OC' 


But  in  the  right  triangles  OB'C  and  OCA\ 

OC  -,  OA'  , 

=  cos  0,  and =  cos  h. 

OB'  OC 

Whence,    cos  c  =  cos  a  cos  6. 


(74) 


100  SPHERICAL   TRlGOXOMETiiY. 

Again,    smA=smB'A'C'  =  ^r^,  =  -jf^,  =  ^r^^,      (75) 

'OB' 

A'C 

and  co^  A  =  Qo&B A C^  = -rr^.  =  ——■.— (76) 

A'B'      AB'      tan  c 

In  like  manner  we  have, 

.     -r,      sin  5  /     \ 

smB  = 5  (77) 

sin  c 

-t  -r-»     Tian  0/  ^     V 

and  cosi3=^ -•  (78) 

tan  c 

140.    From  (75)  and  (76),  we  obtain 

,        J       sin  A      sin  a      tan  c  sin  a 

tan  ^  = = X 


cos  A      sin  c      tan  6      cos  c  tan  6 

Whence  by  (74) , 

,        4  sin  a  tan  a  ,     ^ 

tan^  = —  = (79) 

cos  a  cos  b  tan  6      sin  b 

Similarly, 

,       T)      tan  b  ,      X 

tan^  = iSO) 

sin  a 

141.    Since  sin  a  =  cos  a  tana  (Art.  20),   (75)  may  be 

^^•^t*^^  tana 


.      cos  a  tan  a      tan  c 
sm  ^  = : = 

cos  c  tan  c      cos  c 


cos  a 
Whence  by  (74)  and  (78), 


sin 

Az= 

cos 

B 

COS 

b 

Similarh^, 

sin 

B  = 

COS 

A 

COS  a 


(81) 
(82) 


SPHERICAL   RIGHT   TRIANGLES. 


101 


142.    From  (74),  (8l),  and  (82),  we  have 
cos  c  =  cos  a  cos  h 

_  cos  A      cos  B 
sin  B      sin  A 

=  cot  A  cot  B. 


(83) 


143.  The  proofs  of  Art.  139  cannot  be  regarded  as  gen- 
eral, for  in  the  construction  of  the  figure  we  have  assumed  a 
and  6,  and  therefore  c  and  A  (Art.  138),  to  be  less  than  90°. 

To  prove  formulfe  (74)  to  (73)  universally,  it  is  necessary 
to  consider  two  additional  cases  : 

Case  I.  When  one  of  the  sides  a  and  b  is  <  90°,  and  the 
other  >  90°. 


In  the  right  triangle  ABC,  let  a  be  <  90°  and  b  >  90°. 
Complete  the  lune  ABA'C;  thou  in  the  triangle  A'BC, 

A'B  =  180° -c,  A'  =  A, 

A'C  =  180°  -  b,  A'BC  =  180°  -  B. 

But  by  Art.  138,  c  is  >  90°,  A  <  90°,  and  B  >  90°. 
Hence  each  element,  except  the  right  angle,  of  the  right 
triangle  A'BC  is  <  90°,  and  we  have  by  Art.  139, 

cos  A'B=  cos  a  cos  A'C, 


A ,         sni  a 
sm  A'  = ? 

sin  A'B 

tan  A'C 


cos  A'  = 


At  Tin      Sin  A'C 
sm  A'BC  = » 

sin  A'B 

tan  a 


tan  A'B 


cos  A'BC 


tan  A'B 


102  SPHERICAL   TRIGONOMETRY. 

Putting  for  A'B,  A'C,  A',  and  A'BC  their  values,  we  have 
cos  (180° -c)=  cos  a  cos  (180°-&), 

sin^  = ?ia^ ,      sin(180°-B)  =  ?iBll551=L^, 

sin  (180° -c)  ^  ^      sin  (180° -c) 

cos^^^^"(^^Q°-^),     cos(180°-5)  = ^^^ 

tan  (180° -c)  \  ^      tan(180°-c) 

Whence  by  Art.  47, 

—  cos  c  =  cos  a  ( —  cos  5) , 

.^   A      sin  a                                 .     -D      sin  & 
sin^=: ^  sm5=i ■) 

sin  c  sin  c 

^^^  /f       —  tan  6                                 -D        tan  a 
cos^  = 5  —  eosi5  =  - ; 

—  tan  c  —  tan  c 

and  we  obtain  the  formulae  (74)  to  (78)  as  before. 

In  like  manner,  the  formulae  may  be  proved  in  the  case 
where  a  is  >  90°  and  6  <  90°. 

Case  II.    When  both  a  and  h  are  >  90°. 

a R 


In  the  right  triangle  ABC\  let  a  and  6  be  >  90°. 
Complete  the  lime  ACBC 

By  Art.  138,  c  is  <  90°,  A  >  90°,  and  B  >  90°. 
Hence  each  element,  except  the  right  angle,  of  the  right 
triangle  ABC  is  <  90°,  and  we  have  by  Art.  139, 

cos  c  =  cos  ^C  cos  J5C 

•     T)  int      s'lnBC                     .      ATyrn      sin^C" 
sm^^u'  = 9  sm  ABC  = ? 

sin  c  sin  c 

TyAnt      tan  J.0'                          Annt      tun  BC 
cos  BAC  = —9  cos  ABC   =^ 

tan  c  tan  c 


SPHERICAL    RIGHT   TRIANGLES.  103 

Putting  for  AC\  BC,  BAC,  and  ABC  their  values,  we 
have 

cos  c  =  cos  (180°  -  a)  cos  (180°  -  5) , 

sin(180°-^)  =  ^iHn^5!z:i^,  sin(180°-5)  =  ^Hil^5!zL^, 

sin  c  sin  c 

cos(180°-.l)  =  '^"(''^"°-^-\cos(180°-^)=t""('«^°-'''>. 
tan  c  tan  c 

Whence  by  Art.  47, 

cos  c  =  (  —  COS  a)  (  —  cos  b) , 

^      sin  a                                   .     -n      sin  6 
sm^= 1  sin  5  = ? 

sin  c  sin  c 

.^r,  A      —  t^"  ^                                  r.      —  tan  a 
—  cos  ^  = •)  —  cos  B  = ; 

tan  c  tan  c 

and  we  obtain  the  formuki}  (74)  to  (78)  as  before.  . 


144.    The   forinuLT   of   Arts.    139    to    142    are   collected 
below  for  the  convenience  of  the  student : 

cose  =  cos  a  cos  b. 


i      sin  a 
sin  A  =     — 

•     r>      sin  6 
sill  B  = 

sin  c 

sin  c 

C0Su4=          • 
tanc 

cos  B  = • 

tanc 

tan  A  =  *•'•"«. 

,       n      tan  b 
tan  B  =  — — 

sin  b 

sin  a 

.     ^      cos  B 
sin^= 

siuB-Oos^. 

cos  b 

cos  a 

cosc  = 

cot  A  cot  ^. 

By  comparing  the  formula?  for  the  sines,  cosines,  and  tan- 
gents of  A  and  B  with  the  corresponding  forms  for  plane 
triangles  as  given  in  Arts.  10  and  14,  no  difficulty  will  be 
found  in  retaining  them  in  the  memory. 


104  SPHERICAL   TRIGONOMETRY. 


NAPIER'S  RULES   OF   CIRCULAR  PARTS. 

145.  These  are  two  artificial  rules  which  include  all  the 
formulse  of  the  preceding  article. 

In  any  spherical  right  triangle,  the  elements  a  and  &,  and 
the  complements  of  the  elements  A^  B,  and  c  (written  in 
abbreviated  form,  co.  A,  co.  jB,  and  co.  c),  are  called  the  cir- 
cular' parts. 

CO.  B 


CO. 


If  we  suppose  them  arranged  in  the  order  in  which  the 
letters  occur  in  the  triangle,  any  one  of  the  five  may  be 
taken  and  called  the  middle  par^;  the  two  immediately 
adjacent  are  called  the  adjacent  parts,  and  the  remaining 
two  the  opposite  parts. 

Then  Napier's  rules  are  : 

I.    The  sine  of  the  middle  part  is  equal  to  the  product  of 
the  tangents  of  the  adjacent  p)arts. 

II.  The  sine  of  the  middle  part  is  equal  to  the  product  of 
the  cosines  of  the  opposite  parts. 

146.  Napier's  rules  may  be  proved  by  taking  each  of  the 
circular  parts  in  succession  as  the  middle  part,  and  showing 
that  the  results  agree  with  the  formulae  of  Art.  144. 

1.    If  a  is  the  middle  part,  6  and  qo.B  are  the  adjacent 

parts,   and  co.  c  and  co.  A  the  opposite  parts.      Then  the 

rules  give 

sin  a  —  tan  h  tan  (co.  B) , 

and  sin  a  =  cos  (co.  c)  cos  (co.  A)  ; 

or,  by  Art.  14,    sina  =  tan  b  cotB,   and  sin  a  =  sin  c  sin  ^  ; 

which  agree  with  (80)  and  (75). 


SPHERICAL   RIGHT   TRIANGLES.  105 

2.  If  6  is  the  middle  part,  a  and  co.  A  are  tlie  adjacent 
parts,  and  co.  c  and  co.  B  the  opposite  parts.     Then, 

sin  b  =  tan  a  tan  (co.  ^4) 

=  tana  cot ^4, 
and  sin  b  =  cos  (co.  c)  cos  (co.  B) 

=  sine  sini^ ; 
which  agree  with  (79)  and  (77). 

3.  If  CO.  c  is  the  middle  part,  co.  ^  and  co.  B  are  the 
adjacent  parts,  and  a  and  b  the  opposite  parts.     Then, 

sin  (co.  c)  =  tan  (co.  A)  tan  (co.  B), 

or  cos  c  =  cot  A  cot  B ; 

and      sin  (co.  c)  =  cos  a  cos  6, 

or  cos  c  =  cos  a  cos  b; 

which  agree  with  (83)  and  (74). 

4.  If  CO.  A  is  the  middle  part,  b  and  co.c  are  the  adjacent 
parts,  and  a  and  co.^  the  opposite  parts.     Then, 

sin  (co.  ^4)  =  tan?;  tan  (co.  c) ,    or  cos^l  =  tan  b  cote, 

and     sin  (co.  ^4)  =  cos  a  cos  (co.  B) ,  or  cos^l  =  cos  a  sin  B  ; 

which  agree  with  (76)  and  (82). 

5.  If  CO.  B  is  ti)e  middle  part,  a  and  co.c  are  the  adjacent 
parts,  and  b  and  co.A  the  opposite  parts.     Then, 

sin  ( CO.  J5)  ==  tana  tan  (co.c),    or  cos i^==  tana  cote, 

and     sin  (co.  B)  =  cosb  cos  (co.  A) ,   or  cos^  =  cos 6  sin^l ; 

which  agree  with  (78)  and  (81). 

147.  Writers  on  Trigonometry  differ  as  to  the  practical 
value  of  Napier's  rules  ;  but  in  the  opinion  of  the  highest 
authorities,  it  seems  to  be  regarded  as  preferable  to  attempt 
to  remember  the  formulae  by  comparing  them  with  the  anal- 
ogous forms  for  plane  triangles,  as  stated  in  Art.  144. 


106  SPHERICAL   TRIGONOMETRY. 

SOLUTION    OF    SPHERICAL    RIGHT    TRIANGLES. 

148.  To  solve  a  spherical  right  triangle,  two  elements 
must  be  given  in  addition  to  the  right  angle. 

There  may  be  six  cases  : 

1.  Given  the  hypotenuse  and  an  adjacent  angle. 

2.  Given  an  angle  and  its  opposite  side. 

3.  Given  an  angle  and  its  adjacent  side. 

4.  Given  the  hypotenuse  and  another  side. 

5.  Given  the  tivo  sides  a  and  b. 

6.  Given  the  tivo  angles  A  and  B. 

149.  Either  of  the  above  maj^  be  solved  b}^  aid  of  Art.  144. 
The  formula  for  computing  either  of  the  remaining  elements 

when  any  two  are  given  may  be  found  b}'  the  following  rule  : 

Take  that  forniida  which  involves  the  given  parts  and  the 
required  part. 

If  all  the  remaining  elements  are  required,  the  following 
rule  may  be  found  convenient  in  selecting  the  formulae : 
Take  the  three  formuke  ivhich  involve  the  given  parts. 

150.  It  is  convenient  in  the  solution  to  have  a  check  on 
the  logarithmic  work,  which  maj-  be  done  in  every  case  with- 
out the  necessity  of  looking  out  any  new  logarithms. 

Examples  of  this  will  be  found  in  Art.  153. 
The  check  formula  for  any  particular  case  may  be  selected 
from  the  set  in  Art.  144  by  the  following  rule  : 

Take  that  formula  ivhich  involves  the  three  required  parts. 

Note.  If  Napier's  rules  are  used,  the  following  rule  will  indicate 
which  of  the  circular  parts  corresponding  to  the  given  elements  and  any 
required  element  is  to  be  regarded  as  the  middle  part : 

If  these  three  circular  parts  are  adjacent,  take  the  middle  one  as  the  mid- 
dle part,  and  the  others  are  then  adjacent  parts. 

Ij'  they  are  not  adjacent,  take  the  part  ivhich  is  not  adjacent  to  either  oj 
the  others  as  the  middle  part,  and  the  others  are  then  the  opposite  parts. 

For  the  check  formula,  proceed  as  above  with  the  circular  parts  correspond- 
ing to  the  three  required  elements. 


SPHERICAL  RIGHT   TRIAXGLES.  107 

Thus  if  c  and  A  are  the  given  elements, 

1.  To  find  a,  consider  the  circuhir  parts  a,  co.c,  and  co.^;  of  tliese, 
a  is  the  middle  part,  and  co.c  and  co.^  are  opposite  parts.  Then,  by 
Napier's  rules, 

sin  a  =  cos  (co.c)  cos  (co.^1)  =  sin  c  sin^. 

2.  To  find  h,  tlie  circular  parts  are  b,  co.c,  and  co.A;  in  this  case 
CO. ^  is  the  middle  part,  and  b  and  co.c  are  adjacent  parts.     Then, 

sin  (co.^1)  =  tan  6  tan  (co.c),  or  cos  ^1  =  tan 6  cote. 

3.  To  find  13,  tlie  circular  parts  are  co.B,  co.c,  and  co.^;  co.c  is 
the  middle  part,  and  co.  .1  and  co.  D  are  adjacent  i)arts.     Tlien, 

sin  (co.c)=  tan  (co.  ^1)  tan  (co.  B),  or  cose  =  cot.l  cot^. 

4.  For  the  check  formula,  the  circular  parts  are  a,  b,  and  co.  B;  a  is 
the  middle  part,  and  b  and  co.  B  are  adjacent  parts.     Then, 

sin  a  —  tan  b  tan  (co.  B)  =  tan  b  cot  B. 

151.  lu  solving  spherical  triangles,  cnreful  attention  mnst 
be  paid  to  the  ah/ehraic  signs  of  the  functions  ;  the  cosines, 
tangents,  and  cotangents  of  angles  greater  than  1)0°  being 
taken  negative. 

It  is  convenient  to  place  the  sign  of  each  function  just 
above  or  below  it.  as  illustrated  in  the  examples  of  Art.  lo3  ; 
the  sisfn  of  the  function  in  the  first  member  beino-  then  deter- 
mined  in  accordance  willi  the  i>rinciple  that  like  signs  pro- 
duce -h,  and  unlike  signs  produce  — . 

Note.  In  the  examples  after  tlie  first  of  Art.  153,  the  signs  are 
omitted  in  every  case  wliere  both  factors  of  the  second  member  are  +. 

152.  In  finding  the  angles  corresponding,  if  the  function 
is  a  cosine,  tangent,  or  cotangent,  its  sign  determines  whether 
the  angle  is  less  or  greater  than  1)0°  ;  that  is,  if  it  is  +,  the 
angle  is  <  90°  ;  and  if  it  is  — ,  the  angle  is  >  90°,  and  the 
supplement  of  the  acute  angle  obtained  from  the  tables  must 
be  taken  (Art.  47). 

If  the  function  is  a  sine,  since  the  sine  of  an  angle  is  equal 
to  the  sine  of  its  supplement,  both  the  acnte  angle  obtained 
from  the  tables  and  its  supplement  mnst  be  retained  as  solu- 
tions, unless  the  ambiguit}^  can  be  removed  by  the  principles 
of  Art.  138. 


108  SPHERICAL   TRIGONOMETRY. 

EXAMPLES. 

153.    1.    Given  5=  33°  50',  a  =  108° ;  find  A,  6,  and  c. 

B}'  the  rule  of  Art.  149,  the  formulae  from  Art.  144  are, 

.     ^      cos  A  ^      tan  b  _      tan  a 

sm5= ?      tanij  =  -; — -^      cosB  =  - 

cos  a  sm  a  tan  c 

That  is,  _ 

~.         -        +_      +  ."'"        +^      ~        tanci 

cos  A  =  cos  a  sin  -B,  tan  o  =  sm  a  tan  5,  tan  c  = 


cos-B 

+ 


Hence,  log  cos  ^  =  log  cos  a  +  log  sin  ^, 
log  tan  b  =  log  sin  a  -\-  log  tan  B, 
log  tan  c  =  log  tan  a  —  log  cos  B. 

Since  cos  JL  and  tanc  are  negative,  the  supplements  of  the 
angles  obtained  from  the  tables  most  be  taken  (Art.  152). 

Note.  When  the  supplement  of  the  angle  obtained  from  the  tables 
is  to  be  taken,  it  is  convenient  to  write  180°  minus  the  element  in  the 
first  member,  as  shown  below  in  the  cases  of  A  and  c. 

By  the  rule  of  Art.  150,  the  check  formula  for  this  case  is 

cos  A  =  -^^—^   or   log  cos  A  =  log  tan  b  —  log  tan  c. 
tanc 

The  values  of  log  tan  b  and  log  tan  c  may  be  taken  from 
the  first  part  of  the  work,  and  their  difference  should  be 
equal  to  the  result  previously  found  for  log  cos  ^. 

log  cos  o  =  9.4900  -  10         log  tan  a  =  0.4882 

W  sin5  =  9.7457  -  10         log  cos  B  =  9.9194  -  10 


log  cos  ^=9.2357 -10  log  tan  c  =0.5688 

180°- ^=80°  5.5'  180° -c  =  74°  53.8' 
.-.  ^=99°54.5'.  .-.  c  =105°6.2'. 

log  sin  a  =  9.9782  —  10  Check. 

log  tan  5  =9.8263 -10  log  tan  5  =  9.8045-10 


log  tan  b  =9.8045-10 


log  tan  c  =  0.5688 


b  =  32°  31.1'.  lo^  cos  A  =  9.2357  -  10 


sin 

A  = 

sin  a 

sine 

That 

is, 

SPHERICAL   RIGHT   TRIAXGLES.  109 

2.    Given  c  =  70°  30',  A  =  100°  ;  find  a,  6,  and  B. 
By  Art.  149,  the  three  formulae  are, 

cos  A  =z ,     cos  c  =  cot  A  cot  B. 

tan  c 

-  +        -  -  +        - 

sin  a  =  sin  c  sin  A,   tan  b  =  tan  c  cos  A,   cot  5  =  cos  c  tan  ^. 

The  side  a  is  determined  from  its  sine ;  but  the  ambiguity 
is  removed  by  the  principles  of  Art.  138  ;  for  a  and  A  must 
be  in  the  same  quadrant.  Therefore  a  is  >  90°,  and  the  sup- 
plement of  the  angle  obtained  from  the  table  must  be  taken. 

By  Art.  150,  the  check  formula  is 

tan  B  =  — ,    or   sin  a  =  tan  b  cot  B, 

sin  a 

Note  1.  The  check  formula  should  always  be  expressed  in  terras 
of  the  functions  used  in  determining  the  required  parts;  thus,  in  the 
case  above,  the  check  formula  is  transformed  so  as  to  involve  cot^ 
instead  of  tan  B. 

log  sin  c  =  9.9743  log  cos  c  =  9.5235 

log  sin  A  =  9.9934  log  tan  .1  =  0. 7537 

log  sin  a  =9.9677  log  cot  ^  =  0.2772 

180°-  a  =  68°  10'  180°  -  J5  =  27°  50.6' 

.-.  rt=lll°50'.  .'.  J5  =  152°9.4'. 

log  tan  c  =  0.4509 

log  cos  ^  =  9 . 2  3  9  7  Check. 


log  tan  b  =  9.6906  ^^S  ^^^  ^  =  ^-^^^^ 

lSO°-b  =  26°  7.5'  l^g  ^^^  ^  =  Q-^^72 

.-.  b=  153°  52.5'.        log  sin  a  =  9.9678 

Note  2.  We  observe  here  a  difference  of  .0001  in  the  two  values 
of  log  sin  a.  This  does  not  necessarily  indicate  an  error  in  the  work, 
for  such  a  small  difference  might  easily  be  due  to  the  fact  that  the 
logarithms  are  only  approximate! t/  correct  to  the  fourth  decimal  place. 


110  SPHERICAL   TRIGONOMETRY. 

3.    Given  a  =  132°  6',  b  =  IT  bV  ;  find  A,  B,  and  c. 
In  this  case  the  formulae  are, 

—  ,      tan  a       ,       ^      tan  b         ~~  ""       "^  , 

tan  A  =    .    , ;     tan  B  =  ~. — ?     cos  c  =  cos  a  cos  b. 
sm  b  sm  a 

+ 
The  check  formula  is 

cos  c  =  cot  A  cot  5,    or   cos  c  tan  -4  tan  B=  1. 
That  is,     log  cos  c  +  log  tan  A  +  log  tan  -B  =  log  1  =  0 
log  tan  a  =  0.0440  log  cos  a  =  9.8263 

log  sin  b  =9.9901  log  cos  b  =9.3232 

log  tan  ^  =  0.0539  log  cos  c  =9.1495 

180°  -A  =  48°  32.8'  180°  -  c  =  81°  53.4' 

.-.  ^  =  131°'27.2'.  .-.  c=  98°  6.6'. 

log  tan  b  =0.6670  Check. 

loo-  sin  a  =  9.8704  log  cos  c  =  9.1495 


log  tan  J5  =  0.7966 


loo- tan  ^=0.0539 


log  1  =  0.0000 

4.    Given  A  =  105°  59',  a  =  128°  33' ;  find  6,  5,  and  c. 
The  formulae  are, 


sin  a 

sm  c  = -• 

sm-4 


+ 

sin  b  ■■ 

tana 
tan^ 

+ 
sinjB  = 

cos  J. 

cos  a 

The  check  formula 

is 

sin  5  = 

sin& 

sm  c 


In  this  example,  each  of  the  required  parts  is  determined 
from  its  sine  ;  and  as  the  ambiguit}^  cannot  be  removed  by 
Art.  138,  both  the  acute  angle  obtained  from  the  tables  and 
its  supplement  must  be  retained  in  each  case. 


SPHERICAL  RIGHT   TRIANGLES.  HI 

log  tau  a  =  0.0986  lo^  sin  a  =  9.8932 

l02  tan  A  =  0.5430  looj  sin  .4  =  9.9828 


log  sin  b  =i}Jj3dC)  log  sin  c  =9.9104 
.-.  b  =  -2r3.d',  .-.  c  =  54°  26.7', 

or  158°  56.1'.  or  125°  33.3'. 

log  cos  ^  =  9.4399  Check. 

loor  COS  a  =  9.7946  I02:  sin  b  =  9.5556 


log  sin  B  =  9.6453 


log  sin  c  =9.9104 


.-.  jB=26°13.5',  log  sin  5=9.6452 

or  153°  46.5'. 

It  does  not  follow,  however,  that  these  values  can  be  com- 
bined promiscuously  ;  for  by  Art.  138,  since  a  is  >  90°,  with 
the  vahie  of  b  less  than  90°  must  be  taken  the  value  of  c 
greater  than  90°,  and  the  value  of  B  less  than  90° ;  while 
with  the  value  of  b  greater  than  90°  must  be  taken  the  value 
of  c  less  than  90°,  and  the  value  of  B  greater  than  90°. 

Thus  the  only  solutions  of  the  example  are  : 

1.  ^=21°3.9',       c=  125°  33.3',  15  =26°  13.5'. 

2.  6=  i:)8°56.1',  c  =  54°2(;.7',     i5=  153°  46.5'. 

Note.  The  figure  shows  geometrically  why  there  are  two  solutions 
in  tliis  case. 

B 


For  if  AB  and  J. Care  produced  to  A^,  forming  the  lune  ABA^C, 
the  triangle  A'BC  has  the  side  a  and  the  angle  A^  equal,  respectively, 
to  the  side  a  and  the  angle  A  of  the  triangle  ABC,  and  both  triangles 
are  right-angled  at  C. 

It  is  evident  that  the  sides  A^B  and  A^C  and  the  angle  A^BC  are 
the  supplements  of  the  sides  c  and  h  and  the  angle  ABC,  respectively. 


112  SPHERICAL   TRIGONOMETRY. 

Solve  the  following  spherical  right  triangles  : 

5.  Given  a  =  159%  c  =  137°  20'. 

6.  Given  J.  =    50°  20',  5=  122°  40'. 

7.  Given  a  =  160°,  b  =    38°  30'. 

8.  Given  5=    80°,  b=    67°  40'. 

9.  Given  5  =  112°,  c=    81°  50'. 

10.  Given  a  =61°,  B  =  123°  40  . 

11.  Given  a  =    61°  40',  6  =  144°  10'. 

12.  Given  xl=    99°  50',  a  =  112°. 

13.  Given  b  =    15°,  c  =  152°  20'. 

14.  Given  ^=    62°  59',  B=    37°    4'. 

15.  Given  ^=    73°    7',  c  =  114°32'. 

16.  Given  5=  144°  54',  5  =  146°  32'. 

17.  Given  5=    68°  18',  c=    47°  34'. 

18.  Given  ^=161°  52',  6=131°    8'. 

19.  Given  a  =  113°  25',  6  =  110°  47'. 

20.  Given  a  =  137°    9',  J5=    74°  51'. 

21.  Given  ^=144°  54',  5  =101°  14'. 

22.  Given  a  =    69°  18',  c=    84°  27'. 

SOLUTION  OF  QUADRANTAL  TRIANGLES. 

154.  A  spherical  triangle  is  called  quadrantal  when  it  has 
one  side  equal  to  a  quadrant. 

By  Art.  136,  (/),  the  polar  triangle  of  a  quadi-antal  til- 
angle  is  a  right  triangle. 

Therefore  to  solve  a  quadrantal  triangle  we  have  only  to 
solve  its  polar  triangle,  and  take  the  supplements  of  the  parts 
obtained  by  the  calculation. 

1.  Given  c=90°,  a  =  67°  38',  &  =  48°50';  find  A,  B, 
and  C. 


SPHERICAL  RIGHT   TRIANGLES.  113 

Denoting  the  polar  triangle  bv  A'B'C'^  we  have  by  Art. 
136,  (/)  : 

C  =  90°,  A'  =  112°  22\  B'  =  131°  10' ;   to  find  a',  5',  and  c'. 
By  Art.  144,  the  formula  for  the  solution  are 

~   ,      cos  A'        ~ ,,      cos  B'       '^    ,        ~  i,     ~  -r>, 

cosa'  =  ^ — — ,    cos  6'  = ,,    cosc'=cot^l'cot5'. 

sin  B'  sin  A' 

+  + 

Tlie  check  formula  is      cos  o'  =  cos  a'  cos  6'. 

log  cos  xV  =  0.5804  log  cot  ^'  =  9.6143 

log  sin  JB'  =  9.8767  logcot  5' =  9.9417 


log  cos  (('  =  9.7037  log  cos  c'  =  9.5560 

.-.  180°  -  «'  =  59°  38.2'.  .-.  c'  =  68°  54.8'. 

log  cos  ii'  =  9.8184  Check. 

log  sin.1'  =  9.9660  log  cos  a'  =  9.7037 

log  cos  h'  =  9.8524  l^g  ^'^s  ^'  =  j^f524 

.-.  180° -6' =  44°  36.7'.  log  cos  c'  =9.5561 

Then  in  the  given  quadrantal  triangle,  we  have 
^=180°-a'=    59°  38.2', 
B  =  \so°-b'=    44°  36.7', 
C=180°- c'  =  lll°    5.2'. 

EXAMPLES. 
Solve  the  following  quadrantal  triangles  : 

2.  Given  ^=139°,  6  =  143°. 

3.  Given  ^=    45°  30',  JB=  139°  20'. 

4.  Given  a  =    30°  20',  C  =    42°  40'. 

5.  Given  5=    70°  12',  (7  =  106°  25'. 

6.  Given  A  =  105°  53',   a  =  104°  54'. 


114 


SPHERICAL   TRIGONOMETRY, 


XII.   SPHERICAL  OBLIQUE  TRIANGLES. 

GENERAL  PROPERTIES   OF   SPHERICAL  TRIANGLES. 

155.    In  any  sjjJiericcd  triangle,  the  sines  of  the  sides  are 
proportional  to  the  sines  of  their  opposite  angles* 

G 


Fig.  1. 


-D 


Let  ABC  be  any  spherical  triangle,  and  draw  the  arc  CD 
perpendicular  to  AB. 

There  will  be  two  cases  according  as  CD  falls  upon  AB 
(Fig.  1),  or  upon  ^J5  produced  (Fig.  2). 

In  the  right  triangle  ACD,  in  either  figure,  we  have  by 

Art.  144, 

sin  CD 


sin  A  = 


sin  6 


Also,  in  Fig.  1, 


.     T5      sin  CD 
sin  B  = ? 


sm  a 


and  in  Fig.  2,        sin  B  =  sin  (180°-  CBD) 

=  sin  CBD    (Art.  47) 
sin  CD 


sin  a 


Dividing  these  equations,  we  have  in  either  case 

s'm  CD 
sin  A 


sin  b 


sin  a 


sin  B      sin  CD      sin  b 


(84) 


gm  a 


SPHERICAL   OBLIQUE   TKIAXGLES.  115 


In  like  manner  we  have, 

sin  B  _  sin  b 

sin  C      sin  c 


(85) 


,  sin  C      sin  c  /„_x 

and  -  = (86) 

sin  A      sin  a 

Tiie  above  results  may  be  expressed  more  compactl}'  as 

follows : 

sin  a  _  sin  b  _  sin  c 

sin  A      sin  B      sin  C 

156.  /?i  any  spherical  triangle,  the  cosine  of  either  side  is 
equal  to  the  product  of  the  cosines  of  the  other  two  sides,  plus 
the  continued  p>t'oduct  of  their  sines  and  the  cosine  of  their 
included  angle. 

In  the  right  triangle  BCD,  in  Fig.  1  of  the  preceding 
article,  we  have  by  Art.  Ill, 

cos  a  =  cos  BD  cos  CD 

=  cos  {c- AD)  cos  CD. 

And  in  Fig.  2, 

cos  a  =  cos  BD  cos  CD 

=  cos  (AD  —  c)  cos  CD. 

Whence,  in  either  case, 

cos  a  =  cos  c  cos  AD  cos  CD  -f  sin  c  sin  AD  cos  CD. 

But  in  the  right  triangle  ACD,  by  Art.  144, 

cos  AD  cos  CD  =  cos  b. 

Also, 

cos  b 

sin  AD  cos  CD=  sin  AD -—  =  cos  b  tan  AD 

cos  AD 

=  sin6-^ =  sin  6  cos  J.   (Art.  141). 

tan  6  ^  ^ 

Whence, 

cos  a  =  cos  b  cos  c  +  sin  b  sin  c  cos  A.  (87) 


116  SPHEPJCAL   TRIGONOMETRY. 

Id  like  manner  we  have, 

cos  6  =  cos  c  cos  a  +  sin  c  sin  a  cos  B,  (88) 

and  cos  c  =  cos  a  cos  h  +  sin  a  sin  h  cos  (7.  (89) 

157.    Let  ABC  and  A'B'C  be  a  pair  of  polar  triangles. 


Applying  the  theorem  of  Art.  156  to  the  side  a'  of  the 
triangle  A'B'C ^  we  obtain 

cos  a'  =  cos  b'  cos  c'-\-  sin  b'  sin  c'  cos  -4'. 

Putting  for  a',  6',  c',  and  ^'  their  values  as  given  in  Art. 
136,  (/),  we  have 

cos(180°-^)=cos(180°-5)cos(180°-e) 

+  sin  (180°-  B)  sin  (180°-^)  cos  (180°-  a). 
Whence  by  Art.  47, 

—  cos  J.  =  (—  cos  5)  (— cos(7)  +  sin5sin(7(— cosa). 
That  is, 

cos  A  =  —  cos  B  cos  O  +  sin  ^  sin  C  cos  a.  (90) 

In  like  manner, 

cos  B  =  —  cos  C  cos  A  +  sin  C  sin  A  cos  &,  (91) 

and  cos  C=  —  cos  >4  cos  5  +  sin  ^  sin  B  cos  c.  (92) 

The  above  proof  illustrates  a  very  important  application  of 
the  theory  of  polar  triangles  in  Spherical  Trigonometry^ 

If  any  relation  has  been  found  between  the  elements  of  a 
triangle,  an  analogous  relation  may  be  at  once  derived  from  it, 
in  which  each  side  or  angle  is  replaced  by  the  opposite  angle 
or  side,  with  suitable  modifications  in  the  algebraic  signs. 


SPHERICAL   OBLIQUE   TRIAXGLES.  117 

158.  To  express  the  sines,  cosines,  and  tangents  of  the  half- 
angles  of  a  spherical  triangle  in  terms  of  the  sides  of  the 
triangle. 

From  (87),  Art.  156,  we  obtain 

sin  b  sin  c  cos  A  =  cos  a  —  cos  b  cos  c, 

J      cos  a  —  cos  6  cos  c  .^. 

or,  C0S^  = : — ; — : (BJ 

sm  0  sin  c 

Subtracting  both  members  from  unity, 

^  .      1      cos  a  —  cos  b  cos  c 

1  —  cos  A  =  l 

sin  6  sin  c 

_  cos  b  cos  c  +  sin  6  sin  c  —  cos  a 
sin  b  sin  c 
Whence  b}'  (A),  Art.  75, 

^   .  .,  ,    J      cos  (b  —  c)  —  Gos  a 

2  snr  -kA  = ^^ 

sin  b  sin  c 
But  by  Art.  72, 

cos  y  —  cos  X  =  2  sin  4-  {x  -j-  y)  sin  4-  {x  —  y) . 

Whence, 

2sini[a+(6-0]sini[a-(6-c)] 

Z  sin    7>-xt  =  ; ^ ; ■ 

sni  b  sm  c 

.  ,  ,    .      sin  l{a-\-b  — 

or,  sm-^^  = ^ ~  ,    . 

sm  b  sm  c 

Denoting  a  -|-  6  +  c  by  2  s,  so  that  ,9  is  the  half -sum  of  the 
sides  of  the  triangle,  we  have 

a  +  />  -  c  =  (a  +  &  +  o)  -  2c  =  2s  -  2c  =  2  (.9  -  c) , 
and  a  -  &  +  c  =  (a  +  5  +  c)  -  2  6  =  2  s  —  2  6  =  2  (s  —  6) . 

Whence, 

.  _ .    .      sin  (s  —  b)  sin  (s  —  c) 

sm^ \A— ^^ — .    ,     . — ^ -^ 

sm  0  sm  c 

,  ...         Isin  (s  —  6)  sin  (s  —  c)  ^     . 

and  sm  \A  =  d ^ — .    /    • (93) 

\  smo  sm  c  ^     ^ 


2  1  /I  _  ®^"  T  (<^  +  ^  —  c)  sin  ^(a  —  b-\-c) 


118  SPHERICAL   TRIGONOMETRY. 

In  like  manner  we  have, 


sm^B=  \ 

^sia  (s  —  c)  sin  {s  —  a) 

SI             sin  c  sin  a 

cU.  1  ri  —  > 

I'sin  {s  —  a)  sin  {s  —  &) 

(94) 


and  sin|-C=V ^ — ■ ^^T (95) 

-^  \  sua  a  sin  0 

Again,  adding  both  members  of  (B)  to  unity,  we  have 

cos  a  —  cos  b  cos  c 


1  +  cos  ^  =  1  + 


sin  b  sin  c 

_  cos  a  —  (cos  6  cos  c  —  sin  &  sin  c) 

sin  6  sin  c 

„       2 1   /<      cos  a  —  cos  (5  + c) 

or,  2cos^t^= ^-, — r^ — — 

sm  b  sui  c 

Whence,  as  in  the  proof  of  (93), 

„,    ,      sin -l-(6  H-C  + a)  sin  i- (5 +  c  — a) 

cos^  iA  = — — .    i    .- — -• 

"^  sin  b  sin  c 

But,     6  -f  c  +  a  =  2  s,    and   6  +  c  —  a  =  2  (s  —  a) . 

Whence, 

„  ,    ,      sin  s  sin  (s  —  a) 

cos^  I A  = .    ,     . -1 

^  sin  0  sin  c 


,    ,  sin  s  sin  (s  —  a)  ,     , 

and  cosi^=d ^^-A -•  (96) 

"^  ^      sin  6  sin  c  ^ 


In  like  manner 


,  _  sin  s  sin  (s  —  b)  ,     . 

cos  1-5=  V' ^ ^ -'  (97) 

"^  \       sin  c  sm  a  ^ 


,  ^         sin  s  sm  (s  —  c)  ,     . 

and  cos^C=J -. \    .       •  (98) 

\      sin  a  sm  b 

Dividing  (93)  by  (96) ,  we  obtain 


Isin  (8  — 6)  sin(s  — c)     (      sin  6  sin  c 

tan  ^A  =  V ^    .    I    . sJ- r—. ^ 

"^  \  sm  b  sm  c  \  sin  s  sm  [s—a) 


^J 


sin  (s— &)  sin  (s  — c)  ,     , 

-4 .    ,     ^  '  •  (99) 

sm  s  sin  (s  —  a) 


SPHERICAL   OBLIQUE   TRIANGLES.  119 

In  like  manner, 

,^         Isin  (.s  —  c)  sin  (s  —  a) 

tan  i  ^  =  V ^^ ^/^S:^-^'  (100) 

^  \       sin  s  sm  (8  —  h)  ^       ' 


,  ^,        Isin  (s  —  a)  sin  (s  — 6) 

and  taniC=0 \ —    /,         ,       -  (lOl) 

\       sins  sin  (s  —  c)  ^ 

159.  To  express  the  sines,  cosines,  and  tangents  of  the  half- 
sides  of  a  spherical  triangle  in  terms  of  the  angles  of  the 
triangle. 

From  (90),  Art.  157,  we  obtain 

sin  B  sin  C  cos  a  =  cos  A  -\-  cos  B  cos  (7, 

cos  A  +  cos  B  cos  C  z^x 

or,  cos  a  = ^       . (C) 

sm  B  sm  C 

rr.,          1                     .       cos  yl  +  cos  B  cos  C 
Then,  1  —  cos  a  =  1 ' — ^ 

sin  B  sin  C 

o   .  21          —  (cos  ^  cos  (7— sin  ^  sin  C)  — cos  ^ 
or,  2  sin^  |  a  =  — ^ ^—~  ^—- ^ 

sm  ^  sin  C 

=      cos(^4-<7)  +  cos  J. 
sin  B  sin  O 
Therefore  by  Art.  72, 

2  sin^  ^  a  =  -  2cosi(^4-e+^)cos^(^+0-^) 

sin  B  sin  C 

Denoting  A -\-  B  +  C  by  2  6',  so  that  S  is  the  half-sum  of 
the  angles  of  the  triangle,  we  have  B+C  —  A=2(S  —  A). 

Whence,  smH«=  _<:<^^S  cosjS  -  A)^ 

sin  ^  sin  C 


1  -1  cos >S^  cos  ( 6  —  ^)  /,««N 

and  sin  I  a  =  i  —  ■ ; — ^r^"^ — ^. — "  (102) 

\  sm  B  sin  C 

In  like  manner, 

.    ,7  I     COS  tS  cos  ( aS  —  B)  /^««\ 

and  sin  i c  =    L  COS'S' cos  (5-:C)_  .^^^. 

\  sin  ^4  sin  B 


120  SPHERICAL   TRIGONOMETRY. 

Again,  adding  both  members  of  (C)  to  unity,  we  have 

-,    ,  -,    ,  cos  A  +  cos  B  cos  O 

1  +  cos  a  =  1  H .    ^   .    ^ 

sin  B  sm  C 

_  cos  A  +  cos  B  cos  C  +  sin  B  sin  C 
sin  B  sin  G 


Then,   2cos^ia  =  55^ii+^^5^%l^ 

sin  ^  sin  C 

_2cos|[^+^-0]cosi[^-(^-a)] 
""  sin  B  sin  (7 

_         cosK^  +  ^-C)  cosl-M-^+C) 

or,  cos^  A-  a  .= ^-^ .    -1   .--^ ! ^• 

'  "^  smB  sm  O 

But  ^4-5-C=2(^-0),and^-S  +  (7=2(5'-5). 

Whence 

'  .os2 1  a  -  cos  (.S--^)  cos  (^-C) 

tub   2  t* — :     —    . — — ? 

sm  B  sm  C 

and  eosia  =  J""^<^-:^^'^"^(f-^)-  (105) 

\  sin  B  sin  C 

In  like  manner. 


cos 


1  5  ^  JcQs  (^  -  <^)  cos  (>S  -  ^)  ^  ^^Qg^ 

\  sin  C  sin  ^ 


1^^     /cos(^-^)  COS(/S-^) 
\  sin  A  sin 

Dividing  (102)  by  (105),  we  obtain 


and  cosic  =  .J"""^"~"":^"""^"      ""^-  (107) 

\  sin  A  sin  B 


tania=  J cos  S  cos  (S  -  A)  ^3^ 

^         SI      cos(/^-j5)cos(/S-0)         ^       ^ 

•  In  like  manner. 


and 


tan  ^  5  =  k  - 

cos  ^  cos  (S  —  B) 
coBiS-C)cos{S-Ay 

(109) 

tan  1  c  =  yl  - 

cos  iS  COS  {S  —  C) 

(110) 

oas(S  —  /l^  ons(S—  B) 

SPHERICAL   OBLIQUE   TRIANGLES.  121 

NAPIER'S  ANALOGIES. 
160.    Dividing  (99),  Art.  158,  hy  (100),  we  have 


tanjj4_     /sin  (s  — 6)  sin(s  — c)     I       sin  6*  sin  (s  —  5) 
tan  ^  B      \       sin  §  gin  (s  —  a)        ^  sin  (s  —  c)  sin  (s  —  a) 


sm^AQOs^B_    I  sin"  (.9  —  h)  _  sin(s  —  &) 
cos  ^  ^  sin  4^  i^      V  sin^  (s  —  a)      sin (s  —  o) 

Whence,  by  composition  and  division, 

sin  7}^  cos  j-jB-f-cos  j-xlsin^-B      sin  (g  —  5)  +  sin  (s  —  a) 
sin^^cos^J3  —  cos-^  J.  sin|-^~~  sin  (s  —  b)  —  sin  (s  —  a) 

That  is,  by  Arts.  Go,  6G,  and  73, 

sin  (iA-{-iB)_  tan  j-[.s  -  ?>  +  (g  -  a)] 
sin  (^  J.  —  I--B)  ~  tan  ^[s  —  6  —  (s -^  a)] 

But    s—  ?/  -h  s—  rt  =  2  s  —  (I  —  b  =  c. 

sin4^(^  +  J5)  tan^c 

Hence,  .    T ,  .       J,  =  1 — ,,  ^     ,,•  (ill) 

sin-^(^  — -B)      tanj(a  — 6)  ^ 

161.   Multiplying  (99)  by  (lOO),  we  have 

,    ^        ,^         Isin  (s  — 6)  sin  (s— c)     Isin  (.s— c)sin(6'  — a) 

tanl^tanii3=:  V  — -^^ /      '  ^    ,    \ !^ .    ,    ^  ,,   S 

"^  '^  \      snis  sin(.s  — a)       \      snissm(s— 0) 

sin ^^  sin ^^        |sin-(s  — c)      sin  (s  —  c) 
cos^^cos^-B     V     sin-s      ~      sins 

Whence,  by  composition  and  division, 

cos  ^yl  cos  ^5  —  sin  ^^4  sin  ^B      sin  s  —  sin  {s  —  c) 
cos^yl  (tos^B  -\-  sin-^^^l  sin-^-^  ~  sin  s  -j-  sin  (s  —  c)' 

or,  by  Art.  73,       ^o^i^^  +  ^B)  _  tani[s  -  (.9  -  c)]^ 
-"  cos  (i^ -.1-5)      tani[s +  (s-c)] 

But  s  +  s  —  c=2s  —  c  =  a  +  6. 

TT  cos^(^4-f-jB)  tan^c 

Hence,  yt~a ^  =  z — tt-^t-tt-  (112) 

QO^^{A  —  B)      tan  J(aH- 6)  ^ 


122  SPHERICAL   TRIGONOMETRY. 

162.    Proceeding  as  in  Art.  157,  and  applying  (ill)  to 
the  triangle  A'B'C,  we  obtain, 

sin i(^' -  B')  ~  tan  i(a'-  b') ' 

But,       i(^'+  B')  =  i-(180°-  a  +  180°-  6) 
=  180°-i(a  +  &), 
i{A'-  B')  =  i(180°-  a  -  180°+  6)  =  |(  -a  +  6), 
ic'=i(180°-C)  =  90°-iO, 
and  i(a'-b')=:i(^80°-A-180°-{-B)=i{~'A-\-B). 

Whence, 

sin[180°-i(a  +  5)]_    tan(90°-ia) 
sin-^(  — a  +  6)        "tan^(  — ^  +  5)* 

Therefore,  by  Arts.  42,  46,  and  47, 
sin^(a-|-6)  cot  ^(7 


—  sin  ^{a  —  b)      —tsin^{A  —  B) 

sin-i-(a  +  5)_        cotja 
^^'  sini(a-6)~tani(^-jB)'  ^     ^ 

In  like  manner,  from  (112) ,  we  obtain 

cosj-(^'+^')_       tan|c^ 
cosi(^'-5')~tan.i(a'+6')* 

But,  -|(a'+  b')  =  180°-  i(v4  +  B). 

Whence, 

cos  [180°-  i(a  +  &)]  _         tan  (90°-  j C) 
cos i(-  a  +  &)       "~  tan  [180°-  i(^  +  ^)]' 

Therefore,  by  Arts.  42,  46,  and  47, 

—  cos|-(a  +  6)_  eot^C 
cosi(a  — 5)    ~  —  tan  ^{A  +  -B) ' 

cosi-(a  +  6)  _       cot  JO 
^^'  cosi(a-6)~tani(^  +  ^)'  ^       ^ 


SPHERICAL  OBLIQUE   TRIANGLES.  123 

163.  The  formulae  exemplified  in  Arts.  160,  161,  and  162, 
are  known  as  Napiefs  Analogies.  In  each  case  there  may 
be  other  forms,  according  as  other  elements  are  used. 

SOLUTION   OF   SPHERICAL   OBLIQUE  TRIANGLES. 

164.  In  the  solution  of  spherical  obliqne  triangles,  we 
may  distinguish  six  cases : 

1 .  Given  a  side  and  tivo  adjacent  angles. 

2.  Given  two  sides  and  their  included  angle. 

3.  Given  the  three  sides. 

4.  Given  the  three  angles. 

5.  Given  two  sides  and  the  angle  opposite  to  one  of  them. 
G.  Given  two  angles  and  the  side  opposite  to  one  of  them. 

165.  By  application  of  the  principles  of  Art.  136,  (/), 
the  solution  of  any  example  under  Cases  2,  4,  and  6  may  be 
made  to  depend  upon  the  solution  of  .'mother  example  under 
Cases  1,  3,  and  5,  respectively  ;  and  vice  versa. 

Thus  it  is  not  essential  to  consider  more  than  three  cases 
in  the  solution  of  spherical  oblique  triangles. 

166.  The  student  must  carefully  bear  in  mind  the  remarks 
made  in  Arts.  151  and  152. 

Case  I. 

167.  Given  a  side  and  two  adjacent  angles. 

1.  Given  ^=70°,  5  =  131°  20',  c'=116°;  find  «,  h, 
and  C. 

By  Napier's  Analogies  (Arts.  160,  161),  we  have 

sini(5  +  -'l)  _       tan|c 
sin-2-(5  — ^)      tan  1^(6  — a)' 

and  eosi(^  +  ^)  ^       tan^c 

cos  ^{B  —  A)      tan  |(6  +  a)  * 


124  SPHERICAL   TRIGONOMETRY. 

Whence, 

tan  i(5  —  a)  =  sin  ^-{B  —  A)  esc  ^(B-\-A)  tan  |-c, 
+  _  + 

and     tsin^{b -\-a)  =  cos^{B  — A)  sec^{B-{- A)  tsiuic. 

From  the  data, 

1(5-^)  =  30°  40',  i(^  +  ^)=100°40',  ic  =  58°. 

log  sin  i{B  -A)=  9.7076  log  cos  ^{B-  A)  =  9.9316 

log  CSC  ^{B-\-  A)  =  0.0076         log  sec  i{B  +  A)  =  0.7326 

log  tan|c  =  0.2042  log  tan|-c  =  0.2042 

log  tan  i{b-a)  =9.9194  log  tan  i(6  +  a)  =0.8714 

.-.  i{b  -a)=  39°  42.8'.       180°  -  ^(b  +  a)  =  82°  20.5' 

.-.  i(b-j-a)  =97°  39.5'. 

Then,  a=  i(b-j-  a)  -  ^(b-  a)  =  57°  56.7', 

and  b  =  ^(5+  a)  +  ^{b-  a)  =  137°  22.3'. 

To  find  C,  we  have  by  Art.  162, 

,  _,      sin  4-(6  +«)         .  .^ 

=  sin  |^(& + a)  CSC -^-(^  —  a)  tan  •2-(-S— ^). 
log  sin  i{b  +  a)  =  9.9961 
log  GSGi(b  -  a)  =  0.1946 
log  tan  i{B  -A)=  9.7730 

log  cot  10=9.9637 

.-.  i(7=  47°  23.6',  and  0=  94°  47.2'. 

Note  1.    The  value  of  C  may  also  be  determined  by  the  formula 

Note  2.   The  triangle  is  always  possible  for  any  values  of  the  given 
elements. 


SPHERICAL   OBLIQUE  TRIANGLES.  125 

EXAM  PLES. 
Solve  the  followino;  trianojles  : 

2.  Given  ^=    78°,             B=    41%  c=108°. 

3.  Given  5  =  135°,             0=    50°,  a  =    70°  20'. 

4.  Given  ^=    31°  40',        C'=122°20',  h=    40°  40'. 

5.  Given  J.  =  108°  12',       i^=145°4G',  c=12G°32'. 

Case  II. 
168.    Given  tico  sides  and  their  included  angle. 

1.    Given  b  =  137°  20',  c  =  110°,  .4  =  70°  ;  find  B,  C,  and  a. 

By  Napier's  Analogies  (Art.  1G2), 

sin  ^{b  -\-c)  _        cot  4^  A 
sin  ^(6-0  ~tan^(7i-C')' 

T        cos4-(^  +  c)             cot  I J 
and        '^ — ■ — -  =  —  ^ 

cos  H^  -  c)      tan  i  v-^  +  ^') 

Whence, 

tani(7^-  C)=  s\n  ^{b-c)  csci(&+c)  cotj^, 

-  +  -  + 

and      tan^(Z?+  C)  =  cos^(/>  —  c)  seci(b  -f- c)  cot^^. 

From  the  data, 

^{b-c)  =  10°  40',   |(/j  +  c)  =  12G°40',   4-^4  =  35°. 

From  which  we  find, 

^(B-C)=  18°  14.5',  i  (5  +  C)  =  113°  2.9'. 

Then,      B  =  ^  (B -{- C) -{-^  {B  -  C)=  131°  17.4', 

and  C=^{B-\-C)-^(B-C)=    94°  48.4'. 

To  find  rt,  we  have  by  Art.  IGO, 

tan  i  a  =  2H±i|+^  tan  i  (i  -  0) . 
sm  ^  (is  —  C  ) 

From  which  we  obtain  a  =  57°  56.6'. 


126  SPHERICAL   TRIGONOMETRY. 

Note.   The  triangle  is  always  possible  for  any  values  of  the  given 
elements. 

EXAMPLES. 

Solve  the  following  triangles  : 

2.  Given  a  =    72°,  6=    47°,  C=    33°. 

3.  Given  a  =    98°,  c=    60°,  J5  =  110°. 

4.  Given  5  =  120°  20',       c=    70°  40',  A=    50°. 

5.  Given  a  =125°  10',       6  =  153°  50',  C  =  140°  20'. 

Case  III. 
169.    Given  the  three  sides. 

1.    Givena  =  60°,  5=137°20',  c  =  116°;  find  ^,  £',  and  0. 
By  Art.  158, 

tani^=  jBin(.-5)sin(.;^^ 
^       sin  s  sm  (s  —  a) 

1  •„      .    sin  (s  —  c)  sin  (s— a) 
tan  i  5  =  V^ ^; i— — ^— - — ^5 

^        sin  s  sm  (s—  o) 

tan  *  C=  >i"  («  -  «)  ^i"  (ZEg. 
^       sin  s  sin  (s  —  c) 
From  the  data, 

2s  =  ft  +  6  +  c  =  313°20',  or  s=156°40'. 

AYhence, 

s-a=  96°  40',  s  -  5  =  19°  20',  s  -  c  =  40°  40'. 

log  sin  (s- 6)  =9.5199 

log  sin  (s-c)=  9.8140 

log  CSC  s  =  0.4022 

log  CSC  (s- a)  =0.0029 

2)9.7390 

logtan  1^=9.8695 

0-.  1^  =  36°  31.2',  and  ^=73°  2.4'. 


and 


SPHERICAL   OBLIQUE   TRIANGLES.  127 

In  like  manner  we  find, 

5  =131°  32.2',    and    (7=  96°  55.4'. 

The  values  of  A,  B,  and  C  may  also  be  obtained  by  aid  of 
the  sine  or  cosine  formnlae  of  Art.  158. 

If  all  the  angles  are  to  be  computed,  the  tangent  formulae 
are  the  most  convenient,  as  only  four  different  logarithms  are 
required.  If  but  one  angle  is  required,  the  cosine  formula 
will  be  found  to  involve  the  least  work. 

Note.  The  triangle  is  always  possible  for  any  values  of  the  given 
elements  which  satisfy  the  conditions  of  Art.  138,  (a)  and  (c)  ;  that  is, 
if  a  +  6  +  c  is  <  360°,  and  no  side  is  greater  than  the  sum  of  the  other 
two. 

EXAMPLES. 

Solve  the  following  triangles  : 

2.  Given  a  =    38°,  6  =  51°,  c  =  42°. 

3.  Given  a  =  101°,  b  =  4<J°,  c  =  60°. 

4.  Given  a  =    61°,  />  =  3'.)°,  c  =  92°. 

5.  Given  a  =    62°  20',       ?/=54°10',  c  =  97°50'. 

Case  W. 
170.    Given  the  three  angles. 

1.  Given  ^  =  70°,  5=  131°  10',  0=94°  50';  find  a,  6, 
and  c. 

By  Art.  159, 

tania=v/-        ^os  .S  cos  ( ^  -  ^) 

^     cos  {S  -  B)  cos  {S  -  C) 

f.nn|7>-J  C0S>SC0S(^-^) 

^     COS  {S  -  C)  cos  {S  -  A) 

and  tan  1  c  =  J  cos  ^- cosT^- C ) 

^      cos  {S  -  A)  cos  {S  -  B) 


128  SPHERICAL   TRIGONOMETRY. 

From  the  data, 

S-A=78°,  S-B=16°  50',  S-C=53°10'. 

Note  1.  Since  cos  5  is—  (Art.  37),  while  the  cosines  of  S  —  A, 
S  —  B,  and  S—  C  are  +,  the  quantities  under  the  radical  signs  are 
essentially  positive,  and  hence  no  attention  need  be  paid  to  the  nega- 
tive signs  in  the  formula9. 

log  COS ^S'^  9.9284 

log  cos  {S- A)  =  9.3179 

log  sec  (/S  -  5)  =  0.0190 

log  sec  (^-  C)=  0.2222 

2)9.4875 

logtania=  9.7437 

.•.|a=  28°  59.7',  and  a  =  57°  59.4'. 

In  like  manner  we  find, 

6  =137°  11.8',    and   c=  115°  55.8'. 

The  values  of  a,  5,  and  c  may  also  be  obtained  by  aid  of 
the  sine  or  cosine  formulae  of  Art.  159. 

If  all  the  sides  are  to  be  computed,  the  tangent  formulae 
are  the  most  convenient,  as  only  four  different  logarithms  are 
required.  If  but  one  angle  is  required,  the  sine  formulae  will 
be  found  to  involve  the  least  work. 

Note  2.  The  triangle  is  always  possible  for  any  values  of  the 
given  elements,  provided  S  is  between  90°  and  270°,  and  each  of  the 
quantities  S-A,  S-B,  sind  S  -  C  between  90°  and  -90°  (Art.  37). 

EXAMPLES. 
Solve  the  following  triangles  : 

2.  Given  .4=    75°,  B=    82°,  0=61°. 

3.  Given  ^=120°,  5=130°,  0=80°. 

4.  Given  ^=    91°  10',     B=    85°  40',  0=72°  30'. 

5.  Given  ^=138°  16',     B=    31°  11',  0=35° 53'. 


SPHERICAL   OBLIQUE   TRIANGLES.  129 

Case  Y. 
171.    Given  two  sides  and  the  angle  opposite  to  one  of  them. 

1.    Given  a  =  58°,  6  =  137°  20',  5  =  131' 20' ;  find  A,  (7, 
and  c. 

o      i   *.   1  -  -       sin  ^1      sin  a 
sin  B      sm  h 

or,  sin  ^  =  sin  a  CSC  6  sin  5. 

log  sin  rt  3=9.9284 
log  CSC  h  =  0.1689 
logsini?  =  9.8756 


log  sin^l  =  9.9729 

.-.  .1  =  69° 58',  or  110° 2'  (Art.  152). 
To  find  C  and  c,  we  have  by  Arts.  100  and  102, 

cotiO=^!lii;^ta„K£-^), 

and  tan  |-  c  =  -  — r  ;-^ -. '  tan  1  ( 6  -  a ) , 

sin  ^(5  —  ^)        ^^  ^ 

Using  the  first  value  of  .4,  we  have 

i(i5  +  ^l)=100°39',  ^(i3-.4)=30°41'. 
Also,  ^{h+a)=  97°  40',  ^  (6  -  «)=  39°40'. 
From  which  we  obtain 

C=94°41.6',  and    c=  115° 53. 6'. 
Using  the  second  value  of  ^4,  we  have 

i(i3  +  .4)=120°41',  ^(5 -.4)  =  10° 39'. 
From  w4nch  we  obtain 

(7=  147°  26.4',   and  c  =  150°  56.8'. 
Thus  the  two  solutions  are  : 

1.  A=    69° 58',    C=    94° 41.6',    c=  115° 53.6'. 

2.  ^=110°    2',    (7=147°26.4',    c=  150°  56. 8'. 


130  SPHERICAL    TRIGONOMETRY. 

As  in  the  corresponding  case  in  the  solution  of  plane 
oblique  triangles  (compare  Arts.  124  to  126),  there  may 
sometimes  be  two  solutions,  sometimes  only  one,  and  some- 
times none,  in  an  example  under  Case  V. 

After  the  two  values  of  A  have  been  obtained,  the  number 
of  solutions  may  be  readily  determined  by  inspection  ;  for  by 
Art.  136,  (5),  if  ais  <b,  A  must  be  <  B,  and  if  a  is  >  b, 
A  must  be  >  B. 

That  is,  only  those  values  of  A  can  be  retained  which  are 
greater  or  less  than  B  according  as  a  is  greater  or  less  than  b. 

Thus,  in  Ex.  1,  a  is  given  <  b ;  and  since  both  values  of 
A,  69°  58'  and  110°  2',  are  <  B,  we  have  two  solutions. 

Also  if  the  data  are  such  as  to  make  log  sin  A  positive, 
there  will  be  no  solution  corresponding. 

2.    Given  a  =  58°,  c  =  116°,  0=94°  50';  find  ^,  5,  and  6. 

sin  A      sin  a 


In  this  case, 


sin  C      sin  c 


or,  sin  ^  =  sin  a  esc  c  sin  C. 

log  sin  a  =9.9284 
log  CSC  c  =0.0463 
lo2  sin  0  =  9.9985 


log  sin^=  9.9732 

...  ^==70°  5',  or  109°  55'. 

Since  a  is  given  <  c,  only  values  of  A  which  are  <  O  can 
be  retained  ;  hence  there  is  but  one  solution,  corresponding 
to  the  acute  value  of  A. 

To  find  B  and  6,  we  have  by  Arts,  160  and  162, 

,^      sin-Kc  +  a)-        ,  ,^        .. 
cot  J5  =    ■     I)        4  tan  1(0 -  A) , 

,,       sin  1(0  +  ^),      ,, 
and  tan  2^  =  ^^^^.^  _  ^.t^rii(G- a). 


SPHERICAL  OBLIQUE   TPJAXGLES.  ISl 

Using  the  first  value  of  A,  we  have 

i(C  +  .4)==82°27.o',    ^{C-A)=  12°  22.5'; 
also,         ^(c-\-a)  =  87%  i{c-a)  =  2r. 

From  which  we  obtain 

i5=  131°  21.8',  and  b=  137°  23.6'. 

Thus  the  only  solution  is 

^=70°  5',  ^=131°  21.8',  6  =  137°  23.6'. 

3.    Given  b  =  126°,  c  =  70°,  i3  =  56°  ;  find  C. 

T    .1  .  sin  C      sin  c 

In  this  case,      — = -> 

sin  B      sin  b 

or,  sin  C  =  sin  c  esc  6  sin  5. 

log  sin  c  =  0.0730 
log  CSC  b  =0.0020 
log  sin  ii=  0.0186 

log  sin  (7=0.0836 

.-.  0=74°  20',  or  105°  40'. 

Since  both  values  of  C  are>J5,  while  c  is  given<?>,  there 
is  no  solution. 

EXAMPLES. 
Solve  the  following  triano-les  : 

4.  Given  6  =  99°  40',  c  =  64°  20',  5  =95°  40'. 

5.  Given  a  =  40°,  &  =  118°20',  ^  =  20°  40'. 

6.  Given  a  =115°  20',  c=146°20',  O=141°10'. 

7.  Given  a  =  109°  20',  c  =  82°,  ^=107°  40'. 

8.  Given  6  =  108°  30',  c  =  40°50',  O=39°50'. 

9.  Given  a  =  162°  20',  6  =15°  40',  5=125°. 
10.  Given  a  =  55°,  c  =  138°  10',  A  =  42°  30'. 


132  SPHERICAL   TRIGONOMETKY. 

Case  VI. 

172.    Given  two  angles  and  the  side  opposite  to  one  of  them. 

1.    Given  J.  =  110°,  jB=131°20',  5  =  137°  20';  find  a,  c, 
and  C. 

-r     , ,  .  sin  a      sin  A 

In  this  case,  = •> 

sin  b      sin  B 

or,  sin  a  =  sin  A  esc  B  sin  b. 

log  sin  ^  =  9.9730 
log  csc5=  0.1244 
log  sin  b  =9.8311 

log  sin  a  =  9.9285 

.-.  a  =  58°  1.2',  or  121°  58.8'. 

To  find  c  and  (7,  we  have  by  Arts.  160  and  162, 

^      ,         sm^(B-i-A)  ^      ,.^        . 
tan  -i-c  =  -. — ri^^ aT  tan  f(b  —  a) , 

^.^     sin +(5+ a),       ,,^        .. 
and  cot  i  C  =   .    ,,.        :  tan  ^(B-A). 

sinf(o  — a)  ^  ^ 

Using  the  first  value  of  a,  we  have 

c=  150°  53.8',  and  (7=  147°  23'; 

and  using  the  second  value  of  a, 

c=64°7.8',  and  0  =  85°  17.6'. 
Thus  the  two  solutions  are  : 

1.  a=58°1.2',         c=  150°  53.8',     (7=  147°  23'. 

2.  a  =121°  58.8',     c=64°7.8',         (7=  85°  17.6'. 

In  Case  VI.,  as  well  as  in  Case  V.,  there  are  sometimes 
two  solutions,  sometimes  only  one,  and  sometimes  none ; 
and  it  may  be  shown,  exactly  as  in  Art.  171,  that  only  those 
values  of  a  can  be  retained  tohich  are  greater  or  less  than  b 
according  as  A  is  greater  or  less  than  B. 


SPHERICAL   OBLIQUE   TRIAXGLES.  133 

Also  if  log  sin  a  is  positive,  the  triangle  is  impossible. 

EXAMPLES. 
Solve  the  followinsj  triangles  : 

2.  Given  ^  =  11P/,  (7=80%  c  =  84°. 

3.  Given  yl  =  132°,  ^=140°,  6  =  127°. 

4.  Given  .4  =  62°,  (7=102°,  a  =  64°  30'. 

5.  Given  ^1=133°  50',  73=66°  30',  a  =  81°  10'. 

6.  Given  i5  =  22°  20',  0=146°  40',  c  =  138°  20'. 

7.  Given  .4  =61°  40',  (7=  140°  20',  c  =  150°20'. 

8.  Given  15  =  73°,  (7=  81°  20',  6  =  122°  40'. 

APPLICATIONS. 

173.  In  questions  concerning  geodesy  or  navigation,  the 
earth  may  be  regarded  as  a  sphere. 

Tiie  shorteat  imth  between  any  two  points  is  the  arc  of  a 
great  circle  which  joins  them,  and  the  angles  between  this 
arc  and  the  meridians  of  the  points  determine  the  bearings 
of  the  points  from  each  other. 


Thus,  if  Q  and  Q'  are  the  points,  and  PQ  and  PQ'  their 
meridians,  the  angle  PQQ'  determines  the  bearing  of  Q'  from 
Q,  and  the  angle  PQ'Q  determines  the  bearing  of  Q  from  Q'. 


134  SPHERICAL   TRIGONOMETRY. 

Jf  the  latitudes  and  loDgitudes  of  Q  and  Q'  are  known,  the 
arc  QQ'  and  the  angles  PQQ'  and  PQ'Q  may  be  determined 
by  the  sohition  of  a  spherical  triangle. 

For  if  EE'  is  the  equator,  and  PG  the  meridian  of  Green- 
wich, we  have 

angle  QPQ'=  angle  Q'PG- angle  QPG 

=  longitude  of  Q'—  longitude  of  Q. 

Also,  PQ  =  PE-QE   =  90°- latitude  of  Q, 

and  PQ'=  PE'-^  Q'E'=  90°  + latitude  of  Q'. 

Thus,  in  the  triangle  PQQ',  two  sides  and  their  included 
angle  are  known,  and  the  remaining  elements  ma}'  be  com- 
puted. 

Note.  When  QQ'  has  been  found  in  angular  measure,  its  length  in 
miles  may  be  calculated  by  the  method  of  Art.  132.  In  the  following 
problems  the  diameter  of  the  earth  is  taken  as  7912  miles. 

1.  Boston  lies  in  lat.  42°  21'  N.,  longitude  71°  4'  W.  ;  and 
the  latitude  of  Greenwich  is  51°  29'  N.  Find  the  shortest 
distance  in  miles  between  the  places,  and  the  bearing  of  each 
place  from  the  other. 

2.  Calcutta  lies  in  lat.  22°  33'  N.,  Ion.  88°  19'  E.  ;  and 
Valparaiso  lies  in  lat.  33°  2'  S.,  Ion.  71°  42'  W.  Find  the 
shortest  distance  in  miles  between  the  places,  and  the  bear- 
ing of  each  place  from  the  other. 

3.  Sandy  Hook  lies  in  lat.  40°  28'  N.,  Ion.  74°  1'  W.  ; 
and  Queenstown  lies  in  lat.  51°  50'  N.,  Ion.  8°  19'  W.  In 
what  latitude  does  a  great  circle  course  from  Sandy  Hook  to 
Queenstown  cross  the  meridian  of  50°  W.  ? 


If  the  latitude  of  a  place  is  known,  and  the  altitude  and 
declination  of  the  sun,  the  solution  of  a  spherical  triangle 
serves  to  determine  the  hour  of  the  day  at  the  time  and  place 
of  observation. 


SPHERICAL  OBLIQUE  TRIANGLES.  135 


Thus  let  0  be  the  position  of  the  observer  ;  P  the  celestial 
north  pole  ;  EE'  the  celestial  equator  ;  ////'  the  horizon  ; 
Zthe  zenith  ;  S  the  sun's  position  ;  PSM  a  meridian  passing 
through  the  sun's  position  ;  and  ZSN  a  great  circle  passing 
throngh  Z  and  *S'. 

Then  SM  is  tlie  sun's  declination,  SJSf  its  altitude,  and 
EZ  the  latitude  of  the  plac^e  of  observation. 

Then  in  the  spherical  triangle  SPZ,  we  have 

SP=  PM-  SM=\nf-  the  sun's  declination, 
SZ=  ZN-  6'.Y=-  1)0"- the  sun's  altitude, 
and      PZ=  EP  -  EZ  =  'J0°-  the  latitude  of  the  place. 

That  is,  the  three  sides  of  the  triangle  SPZ  are  known, 
and  the  angle  SPZ  may  be  computed. 

If  24  hours  is  multiplied  by  the  ratio  of  this  angle  to  360°, 
we  have  the  time  required  for  the  sun  to  move  from  S  to  the 
meridian  EP. 

Hence,  if  this  time  is  subtracted  from  12  o'clock,  if  the 
observation  is  made  in  the  morning,  or  added,  if  made  in  the 
afternoon,  we  obtain  the  hour  of  the  da}'  at  the  time  and 
place  of  observation. 

If  the  Greenwich  time  of  the  observation  is  noted  on  a 
chronometer,  the  difference  between  this  and  the  local  time 
as  calculated  above  serves  to  determine  the  longitude  of  the 
place  of  observation. 


136  SPHERICAL   TRIGON^OMETRY. 

In  reducing  time  to  longitude,  it  should  be  borne  in  mind 
that  24  hours  of  time  correspond  to  360°  of  longitude;  that 
is,  one  hour  of  time  corresponds  to  15°  of  longitude,  one 
minute  to  15',  and  one  second  to  15''. 

4.  A  mariner  observes  the  altitude  of  the  sun  to  be 
14°  18',  its  decUnation  beiug  18°  36'  N.  If  the  latitude  of  the 
vessel  is  50°  13'  N.,  and  the  observation  is  made  in  the  morn- 
ing, find  the  hour  of  the  day.  If  the  observation  is  taken 
at  9  A.M.,  Greenwich  time,  what  is  the  longitude  of  the  vessel? 

5.  What  will  be  the  altitude  of  the  sun  at  4  p.m.  in  San 
Francisco,  lat.  37°  48'  N.,  its  declination  being  12°  S.? 

6.  In  Melbourne,  lat.  37°  49'  S.,  the  altitude  of  the  sun 
is  observed  to  be  25°  46'.  If  the  sun's  declination  is  3°  S., 
and  the  observation  is  made  in  the  morning,  find  the  hour  of 
the  day. 

7.  At  what  hour  will  the  sun  rise  in  Boston,  lat.  42°  21'  N., 
when  its  declination  is  15°  N.  ? 

Note.  At  sunrise  the  sun's  altitude  is  0,  so  that  the  arc  SZ  be- 
comes 90°. 


FORMULA. 


137 


FORMULA. 


PLANE  TRIGONOMETRY. 


Art.  10. 

sm^l  =  --) 
c 

tan  A  =  -1 
b 

sec^  =  -1 
b 

A         ^ 

cos^  =  -, 

c 

cot  A  =  -1 
a 

A           ^ 

CSC  A  =  -' 

a 

sin  jB  =  - , 

c 

tanB  =  -? 
a 

secB  =  -1 

a 

cos  B  =  -■) 

c 

cot  B  =-^ 

b 

CSC  5=-. 
b 

Art.  18. 

sin^ 

1         . 

= 7?    tan 

.          1 
A  = •>   sec 

A-     1     , 

(1) 


(2) 


CSC  A 


cot^ 


COS  ^1 

_1 

sin^ 


cos^  = ?  cot^l  = ,  csc^l  = 

sec  A  tan  A 

Art.  19.  sin^  A  +  cos'  .1=1. 

Art.  20. 

,        J       sin  A  ,^^  .    .      cos^ 

tan  A  = 7-  (5)  cot  A 


sin^l 
csc-^=  1  +cot-^. 


cos  A 
Art.  21. 

sec2^=lH-tanM.  (7) 
Art.  42. 

sin  (— ^)  =  — sin^,  cos(— .4)=      cos^, 

tan  (—.4)= —tan  J.,         cot  (—^)  = —cot  J., 
sec(  — ^)=      sec^,         csc(— ^)=  — csc^. 

Art.  44. 

sin  (90°  +  ^)=      cosu4,  cos  (90°+^)= -sin^,  "] 

tan  (90°+ 4)= -cot  ^,  cot  (90°+^)  = -tan ^,   f 

sec  (90°+^)= -CSC  ^,  csc(90°+^)=      secX  J 


(3) 

W 
(6) 

(8) 
(9) 


(10) 


138  FORMULA. 

Art.  65.     sin  (x-\-y)  =  sin  x  cos  y  +  cos  x  sin  y.  (ll) 

cos  {x  -\-y)  =  cos  X  cos  y  —  sin  x  sin  2/.  (12) 

Art.  66.     sin  (a;  —  2/)  =  sin  ic  cos  2/ —  cos  a;  sin  2/.  (13) 

cos  (ic  —  2/)  =  cos  X  cos  2/  +  sin  x  sin  2/.  (14) 

tan  a;  +  tan  V  /__x 

Art.  71.     tan(^  +  y)=^-^^^^.  (15) 

tan  X  —  tan  2/  /,  g\ 
^        '^^~  1 -}-t&.nxtsiny 

cot  a;  cot  2/  —  1  /^  n\ 

^  cot  2/  + cot  X 

^,          -      cota;cot2/  +  l  /,«>. 

cot  (.T  —  2/)  =  — 7 ; —  CJ-8) 

^  cot  2/  —  cot  X 

Art.  72.  sin  a;  +  sin  2/=  2sin  ^(a;  +  2/)  cos|-(cc— 2/).  (19) 
sin  x  —  siny=z  2  cos  ^{x  +  v/)  sin  v;(x—y) .  (20) 
cosaj  +  cos2/=  2cos|-(a;H-2y)  cos-^(a;— 2/).  (21) 
cos  a;  — cos 2/=  —2  sin  i(aj  +  y)  sin  i(^— 2/)-  (22) 

Art.  73.  sin  x  +  sin  2/  tan  ^{x-\-y)^  .^^. 
sin  ic  —  sin  y  ~~  tan  ^  {x  —  2/) 

Art.  74.             sin  2. ^•  =  2  sin  a?  cos  a;.  (24) 

cos  2  a;  =  cos^  a?  —  sin^  x.  (25) 

cos  2  a.' =  1  -2sin2a;.  (26) 

cos  2  a;  =  2  cos^  a;  —  1.  (27) 

.      J.            2  tan  a;  /-o\ 

tan  2  a;  = —  (28) 

1  —  tan-"  X 

C0t2a;=:^^^'^~^-  (29) 
2  cot  a; 


FORMULA.  139 


Art.  75. 


sinia;=^ (30)      cos|-a;  =  %|--^— (31) 

tsini-x  =  \l (32) 

tan*a;= (33)      taDia;  = (34) 

1  +  cos  X  sin  X 

.  ,          1  4- cos  a;          sin  x  ^__x 

cot^a;  =  — = (35) 

sin  X  1  —  cos  X 

Art.  111. 

4/r=c-sin2^.                 (36)      4/r=c-sin25.  (37) 

2K=d-Q0iA.                  (38)      2  K=b- cot  B.  (39) 

2/f=a-tanJ5.  (40)  2A^=?;-tan^.  .  (41) 
2K=a^(c  +  a){c-a).  (42)  2  A^=  ?>  V(c-f /j)  (c-6).  (43) 
2K=ab.                           (44) 

Art.  113. 

^^  =  ±     (45)              ?i^=?'.     (46)              5iB^=l  (47) 

slnZ^      6     ^     ^            sinC      c     ^     ^            sin^      a  ^     ^ 

Art.  114.        cj±b^tani(A  +  B)^  ^^3^ 
a-b      tain  {A -B) 


6  4-c^tau^(J3+0) 
6-c~tani(J5-(7)' 

c  +  <x_tan^(C  +  ^) 
c  — a      tan|((7  — ^) 


(49) 
(50) 


Art.  115.         ^±^  =         ^^^y^^  .  (51) 

a-b      tan^^A-B)  ^     ^ 

Art.  116.  a-  =  b-  -{-cr  -2bc  cos  A  (52) 

62  ^  (.2  ^  ^^2  _  2  ca  cos  ^.  (53) 

(^  =  cr  +  b'  -2ab  cos  O.  (54) 


140  FORMULA. 

Art.  117.        cosA  =  ^^^^^^^'  (55) 

2  be 

cos.B^^  +  "'-^'-  (56) 

2ca 

cosC="'  +  ^'-'^-  (57) 

2  ah 


Art.  118.     smi^=J("~^l^"~"^»  (58) 

^  he 


siniiJ=Jii^£Hi^l^.  (59) 

\  ca 

siniC?=J5E£)5^.  (60) 

cosi^=Ji(^.  (61) 

cosiB=J?i^;=S-  (62) 

\       ca 

cosiC=J^5E£).  (63) 

\       ab 


^^,^^=M±zm^.  (64) 


Art.  119. 


tani^==J^^-;>^-\--">.  (65) 

\       s{s  —  h) 

tan  iC=JBi£Hl.  (66) 

^  \       s  (s  —  c) 


2^=5csin^.  (67)  2^= -, — (70) 

2K=casmB.  (68)  „     h^  sinC smA  ,__n 

ZK  = : — vvi; 

2^=a6sina.  (69)  ^   •     a    -    -n 

^j^^c'smAsmB^  (72) 

sinO 


K=>Js{s-a){s-h){s-c).  (73) 


FORMULA.  141 

SPHERICAL    TRIGONOMETRY. 
Art.  139. 

cos  c  =  COS  a  cos  b.  (74) 

4      sin  a  .„^s  .     ^      sin  6  ^     ^ 

smA  =  - (75)  sin^  =  ^ (77) 

sine  smc  ^     ^ 

A      tan  6  /„^x  ^      tan  ft  ,     . 

COS^  = ■'  (76)  C0S5  =  7 -•  (78) 

tanc  tanc  ^ 

Art.  140. 

tan^  =  ^-^^.  (79)  tan5  =  *^.  (80) 

sin  b  sin  a 

Art.  141. 

J      cos  B  /„,  X  .     -D      cos^  /__K 

sin  A  = (81)  sm  B  = (82) 

cos  6  cos  a 

Art.  142. 

cos  c  =  cot  A  cot  B.  (83) 

Art.  155.  §^  =  ^-  (84) 

(85) 

...  (86) 

Sin  A      sin  a 

Art.    156. 

cos  ct  =  COS  b  cos  c  +  sin  b  sin  c  cos.^^  (87) 

cos  b  =  cos  c  cos  a  +  sin  c  sintt  cos  5.  (88) 

cos  c  =  cos  a  cos  b  +  sin  a  sin  b  cos  C.  (89) 

Art.  157. 

cos  A=  —  cos  B  cos  (7  +  sin  5  sin  O  cos  a.  (90) 

cos B=  —  cos  C  cos  ^  +  sin  C sin ^  cos  b.  (91) 

cos  (7=  —cos  A  cos  5  +  sin  A  sin  5  cos  c.  (92) 


sin^ 

sin  6 

sin  B 

sin  6 

sin  C 

sine 

sin  C 

sine 

142  FORMULA. 

.     ,    ,         /sin(s  —  6)  sin(s  —  c)  ^.-n 

Art.  158.     sini^=v— ^ — .    '    • (^3) 

^  \  sin  0  sm  c 

^.  lsm(«-c)sin(.-g_  (3^) 

^  \  sm  c  sm  (X 

sin  *  O  =  /'"(«-'^)^'"(«-J).  (95) 

\  sm  a  sm  o 


cosi^=>'°^.t^^~'^^-  (96) 

"=  \       sm  0  sm  c 


"^  \      sm  c  sm  a 

,^        lsmssin(s  — c)  /qqx 

cos  1^(7=.    : \     ,     '-  (98) 

\       sm  a  sm  o 

tan  i ^  =  , Ni"  («-?;)  sin  (a -^.  (99) 

\       sm  s  sm  (s  —  a) 

tan  J  B  =  , p(>-c)  sin  (.-_«).  ^^^^^ 

\       sm  s  sm  (s  —  o) 

tan  1 0  =  ,  |sin(«-")^i"(«^.  (101) 

\        sms  sm  (s—c) 

A-i.     -icn  -1  I       cos /S  cos  (/S' —  ^)  /inoN 

Art.  159.     sm4-a=. -, — — ^ — - — ^-  (102) 

\  sm  B  sm  (J 

.      1,  I       cos  >i^  cos  (/S  —  5)  /ino\ 

\  sm  C  sm  A 

.       1  I       COSTS'  cos  (S—  C)  /TnA\ 

^^■'*^=V sin ^  sin .B      '  ^'°*^ 

eos  i a  =    looHS-B)oos(S^Vl_   ^      (,05^ 
\  sm  B  sm  G 


1,  IcOS  (O  —  U ■)  COS  ^O  —  ^)  /TrtCN 

COS  i  6  =  i  ^ ■    ri  -     A (^°^) 

\  sm  C  sm  ^ 


'cos(^-C)cos(^^-^) 


FORMULA.  143 


"^  \  smA  sin  B 

tan^a-y^      cos  (6' -5)  cos  (.S  -  O)         ^       ^ 


tan  i  6  =  J ^^^(^Z:^.        (109) 

^  \      cos(^— C)  cos(aS-^) 

tan  i  c  =  J-       ^^f^^l^S^CJ^         (,,0) 
"^  \      COS  {S  —  A)  cos  (>S^  —  B) 

Art.  160. 

sin  ^  (^4  +  ^)  _       tan  j^  c  (111) 

sm^(^-^)      tan  ^  (a -6)* 


Art.  161. 


cos^(.l-ii)      tan  ^  (a +  6) 


Art.  162. 

sin  i-  (ff  +  ft)  ^        cot  i  (7 
sin  i  (a -ft)      tan^(^-J5)' 

cos  4^  (g  H-  ft)  _        cot  ^  (7 
cos  i(a-b)      tan  i  (^  +  5) 


(113) 
(114) 


AI^SWEES 


Art.  9;  page  3. 

9.  28°  38' 52.4".  14.  114°  35' 29.6". 

13.  42°  58'  18.6".  15.  100°  54'  5.1". 
16.  30°  40'  33.8". 

Art.  91;  page  57. 

2.  .7781.     7.  1.3222.  12.  1.9912.  17.  2.1303. 

3.  1.1461.    8.  1.7993.  13.  2.0212.  18.  2.2252. 

4.  .9030.     9.  1.7481.  14.  2.0491.  19.  2.1673. 

5.  1.0791.    10.  1.9242.  15.  2.1582.  20.  2. 5741. 

6.  1.1761.    11.  1.6532.  16.  2.3343.  21.  2.5353. 

Art.  93;  page  58. 

2.  .3680.     5.  1.5441.  8.  .2252.  11.  .8539. 

3.  .1549.     6.  .1182.  9.  2.2431.  12.  .7660. 

4.  .5229.     7.  2.0970.  10.  1.0458.  13.  .7360. 

Art.  96;  page  59. 

3.  .2863.     9.  4.5844.  15.  .1165.  22.  .2601. 

4.  2.7090.    10.  3.2620.  16.  .3860.  23.  .6884. 

5.  4.2255.    11.  .9801.  17.  .2212.  24.  .1840. 

6.  .1398.    12.  .4225.  18.  .1750.  25.  .2215. 

7.  .7194.    13.  .1590.  20.  2.6145.  26.  .2494. 

8.  .6611.    14.  .0430.  21.  .1678.  27.  .1449. 


ANSWERS. 

14£ 

Art. 

98;   page  60. 

2. 

.2552. 

7. 

7.7323-10. 

12. 

2.4804. 

3. 

.3522. 

8. 

6.4983-10. 

13. 

8.7905-10 

4. 

9.2922-10. 

9. 

3.8663. 

14. 

6.3588. 

5. 

8.6811-10. 

10. 

.6074. 

15. 

.1964. 

6. 

1.5841. 

11. 

9.6511-10. 

16. 

.1688. 

Art. 

105: 

;   pages  65  and 

66. 

1. 

8.454. 

19. 

-1.184. 

40. 

.5010. 

2. 

10.73. 

20. 

.000007038. 

41. 

1.062. 

3. 

—  2202. 

21. 

2.924. 

42. 

-.9102. 

4. 

.2179. 

22. 

.9146. 

43. 

1.093. 

5. 

.01157. 

23. 

4.638. 

44. 

.7035.' 

6. 

-.7032. 

24. 

.0000639. 

45. 

.5807. 

7. 

7.G72. 

25. 

1.414. 

46. 

-.6313. 

8. 

.6688. 

26. 

1.495. 

47. 

24.62. 

9. 

-3.908. 

27. 

-1.246. 

48. 

.2979. 

10. 

1782. 

28. 

.6553. 

49. 

98.50. 

11. 

.3500. 

29. 

.2846. 

50. 

1.660. 

12. 

-  .4748. 

30. 

2.372. 

51. 

3.076. 

13. 

.4127. 

31. 

-.5142. 

52. 

.8678. 

14. 

-4.671. 

32. 

.1588. 

53. 

1.134. 

15. 

.2415. 

35. 

5.883. 

54. 

.5881. 

16. 

-.0725. 

36. 

.7885. 

55. 

1.805. 

17. 

13587. 

37. 

1.195. 

56. 

.003229. 

18. 

.006415. 

38. 
39. 

.6803. 
.6443. 

57. 

.03344. 

Art.  110;   pages  69  to  72. 

1.  a  =  7.708,    5  =  8.124.         3.    a  =  24.37,         c  =  53.56. 

2.  6  =  1883,     c  =  2019.5.       4.  ^  =  43°  17.9',   6  =  .6622. 


146  ANSWERS. 

5.  ^=68°  12.2',  c  =  5.385. 

6.  b  =26.91,  c  =  87.64. 

7.  a  =.02309,  6  =  .01452.  31.    c  =  25.206. 

8.  ^  =  55°  36.1',  a  =  4.216.  32.  a  =1.735. 

9.  a  =5571,  c=7007.  33.    c  =  2725.6. 

10.  A  =  4:1°  2  A',  c  =  153.8.  34.  ^  =  47°  42.9'. 

11.  b  =167.6,  c  =  230.5.  35.  a  =.8346. 

12.  a  =30.51,  5  =  18.59.  36.  a  =49.25. 

13.  ^  =  47°  52.5',  6  =  184.7.  37.   14.106  in. 

14.  a  =2.847,  c  =  3.287.  38.   78.12ft. 

15.  a  =5.4125,  c=14.306.  39.    31°  47.1'. 

16.  a  =.7133,  5  =.1367.  40.    36°  37.9'. 

17.  ^=51°  51.9',  c  =  811.7.  41.    99.45  miles. 
18.6=42.27,  c=208.15.  42.    11.371  in.    "^ 

19.  ^  =  58°  35.7',  a  =  .0409.  43.    19°  50'. 

20.  6=76.13,  c=1877.  44.    399.5ft. 

21.  ^  =  36°  45.9',  c  =  41.22.  45.    First,  25.22  miles; 

22.  A  =  65°  30',  a  =  3.153.  second,  30.07  miles. 

23.  a  =36992,  6  =  4021.  46.    21.65  ft. 

24.  a  =410.5,  c  =  456.7.  47.    14.487 in.  ;  15.682  in. 

25.  ^  =  55°  44.1',  c  =  411.5.  48.    453.7  ft. 

26.  ^  =  58°  40',  6=1.0405.  49.    17.26  in. 

27.  6  =.3245,  c=.8828.  50.    389.4ft. 

28.  a  =264.9,  6=75.95.  51.    437.6. 

29.  a  =176.64,  c  =  213.65.  52.    5.773  in. 

30.  ^  =  30°  17.2',  c  =  20.04.  53.    481.9  ft. 

54.   Eate,  6.792  miles  an  hour;  bearing,  63°  8.4'  west  of 
north. 

Art.  112  ;  page  74. 

2.  69.03.  5.    .08938.  8.    1.223. 

3.  .151.  6.    8208.  9.    107.2. 

4.  5695.  7.    .002245.  10.    .1017. 


ANSWERS. 


147 


Art.  121 ;  page  84. 
2.    5=15.837,  c  =  14.703.         7.    a  =  .011162, 5=.006962. 


3.  a  =.445,       c=.9942. 

4.  a=. 01913,  ?;=. 02272. 

5.  rt=  33.78,    c  =  18.54. 

6.  6  =  8.24,      c  =  5.464. 


8.  6  =  447.4,    c  =  425.7. 

9.  a  =  342.6,    c  =  303.3. 
10.    a  =11.067,  6=6.067. 


Art.  122;  page  85. 


2.  ^  =  100^56.7',  6  =  19.78. 

3.  A=   83°  14.7',  c  =  383.5. 

4.  B=   39°  15.8',  a  =  3. 211. 

5.  A=   24°41.8',  c=.5886. 

6.  J5=143°29.1',  a  =  886.4. 


7.  ^=35°  1.9',     6=12.78. 

8.  ^=59°  16.3',   c=60.74. 

9.  7^  =  21°  1.2',     a=.06699. 
10.  .4=40°52' 19", 6=77.14. 


Art.  123;  page  88. 

3.  yl=28°57.4',  B=    46°  34.2',  C=  104°  28.4'. 

4.  J  =  34°  46.4',  B=    86°  24.8',  C=  58°  49'. 

5.  .4  =74°  40',      B=    47°  46.4',  C=  57°  33.4'. 

6.  .4  =  58°  26',      B=    74°  23.8',  C=  47°  10.6'. 

7.  .4  =  55°  55.4',  B=    79°  43.8',  C=  44°  20'. 

8.  .4  =  49°  24.2',  B=    58°  38',  C=  71°  57.8'. 

9.  .4  =  60°  50.8',  B=    46°  6.2',  (7=  73°  1.6'. 
10.  ^=18°  12.4',  5  =135°  50.6',  C=  25°  56.6'. 


Art.  127;  pages  91  and  92. 

6.  B=    39°  21.3',  c  =  5.511. 

7.  J5=    33°  28.4',  a  =118.33; 
or,  5  =146°  31.6',  a  =14.58. 

8.  0=    53°  18.9',  a  =.07508. 

9.  A=    19°  18.1',  6  =  2.522. 


148  ANSWERS. 

10.  A=    79°  20',  G=. 11416; 
or,  ^  =  100°  40',  c=. 05121. 

11.  Impossible. 

12.  B=    24°  5.4',  a  =103.3. 

13.  0=    90°,  &  =  5.007. 

14.  C=    67°10',  a  =  6.918; 
or,  0=112°  50',  a  =  2.913. 

15.  A=    46°  53.3',  c=  141.48. 

16.  Impossible. 

17.  0=    24°  31.4',  &  =  1.0637. 

18.  ^=    90°,  c=  127.32. 

19.  A=    70°  12',  6  =  .2879; 
or,  ^=109°  48',  &=.1045. 

20.  C=    37°  10',  a  =  289.3. 

Art.  128 ;  page  93. 

2.  1077.9.  6.    2604.  10.  .1273. 

3.  14.697.  7.    1.353.  11.  4491. 

4.  3.257.  8.    .06858.  12.  .00001817. 

5.  5114.  9.    215.9.  13.  11.732. 

Art.  129 ;  pages  93  and  94. 

1.  Height  of  tower,  153.64  ft.  ;  distance  from  first  point, 
117.27  ft.  ;  from  second,  217.27  ft. 

2.  29800  square  rods. 

3.  247.7  ft. 

4.  4.927  miles. 

5.  From  first  position,  9.9  miles  ;    from  second,  19.122 
miles. 

6.  298  ft. 

7.  Distance,  91.66  ft.  ;  height,  33.9  ft. 

8.  1113.1  ft.  9.    248  ft.  10.    1569.6. 


ANSWERS.  149 

Art.  153;    page  112. 

5.  ^=148°  5',          B=    65°  23.2',  6=    38°  2'. 

6.  a  =    40°  41.8',     5  ==  134°  30.8',  c  =  122°  7.5'. 

7.  J.  =  140°  41',         S=    66°  43.8',  c  =  137°  20'. 

8.  ^=    27°  11.6',     a=    25°  25.2',  c=    69°  54' ; 
or,  ^=152°  48.4',     a  =  154°  34.8',  c=110°6'. 
9.' ^  =  109°  22.5',     a  =  110°  57.6',  5  =113°  22'. 

10.  ^=    66°  12.1',     ?>  =127°  17.4',  c  =  107°  5.1'. 

11.  .1=    72°  28.9',     ^=140°  38.1',  c  =  112°  37.7'. 

12.  B=    '2ri.2\       h=    25°  24.4',  c  =  109°  46' ; 
or,  5  =  152°  52.8',     6  =  154°  35.6',  c  =    70°  14'. 

13.  .4=  120°  44.3',     a  =  156°  30',  J3  =    33°  52.6'. 

14.  a=    41°  5.5',       h  =    26°  25',  c=    47°  32.1'. 

15.  a=    60°  31.4',     2^  =143°  50',  6  =147°  32.1'. 

16.  .1=    78°  46.7',     a=    70°  10',  c  =  106°  27.5'; 
or,  ^1  =  101°  13.3',     «  =  109°50',  c=    73°  32.5'. 

17.  .1=    30°  32.1',     a=    22°  1.1',  h=    43°  17.9'. 

18.  a=166°8.7',       5=101°  48.9',  c=    50°  18.4'. 

19.  ^1=112°  2.5',       i5=109°12',  c=    81°  53.6'. 

20.  yl=135°2.5',       h=    68°17.3',  c  =  105°  44'. 

21.  a  =146°  32.5',     Z.  =  109°  48.1',  c=    73°  35'. 

22.  A=    70°,              i^=    75°  6.2',  6=    74°  7.3'. 

Art.  154;    page  113. 

2.  a=117°1.2',       J5=153°41.9',  C=132°34'. 

3.  a  =    57°  22.1',     6  =129°  41.6',  0=57°  52.5'. 

4.  A=    20°  0.9',       5=  141°  29.6',  6  =113°  17.1'. 

5.  ^=    33°  27.5',     a=    35°  4.4',  6=   78°  46.7'. 

6.  5=    69°  16',        5=    70°,  0=   84°  30' ; 
or,  5  =  110°  44',        h  =110°,  C=   95°  30'. 


150  ANSWERS. 

Art.  167;    page  125. 

2.  a=    95°  37.8',      6=    41°  52.2',  (7=  110°  48.8'. 

3.  6  =  120°  16.6',      c=    69°  19.6',  A=    50°  26.2'. 

4.  a=    34°  3',  c=    64°  19.4',  B=    37°  39.6'. 

5.  a=    69°  4',  6  =  146°  25.6',  0=    125°  12.2'. 

Art.  168;   page  126. 

2.  ^  =  121°  32.8',     J5=    40°  56.8',  c=    37°  25.8'. 

3.  ^=    86°59.7',     C=    60°50.9',  &  =  111°17'. 

4.  5  =134°  57.3',     0=    50°  41.1',  a=    69°  8.8'. ' 

5.  ^  =  147°  29',        5=  163°  8.6',  c=    76°  8.4'. 

Art.  169;   page  127. 

2.  A=    51°  58.2',     B=    83°  54.4',  C=    58°  53.2'. 

3.  ^=142°  32.8',     B=    27°  52.6',  (7=    32°  27.2'. 

4.  J.=    35°  31',        5=    24°  42.6',  0=138°  24.8'. 

6.  A=    47°  21.2',     5=    42°  19.4',  0=124°  38'. 

Art.  170;  page  128. 

2.  a=    67°  51.8',      6=    71°  44.4',  c=    57°. 

3.  a  =  144°  9.8',        6  =  148°  48.6',  c=    41°  43.6'. 

4.  a=    89°  51.2',      b=    85°  49.2',  c=    72°  31.8'. 

5.  a  =100°  4.6',        5=    49°  58.8',  c=    60°  6'. 

Art.  171;    page  131. 

4.  0=    65°  30',        A=    97°  20',  a  =  100°  44.6'. 

5.  B=    42°  40',         0  =  159°  54',  c  =  153°  30.2'; 
or,  5  =  137°  20',        0=    50°  20.6',  c=    90°  9.6'. 

6.  Impossible. 

7.  0=    90°,  ^  =  113°  36.9',  6  =  114°  51.9'. 


ANSWERS.  151 

8.  ^=    68°  18',        .4  =132°  33.8',      a  =131°  15.8'; 
or,  5  =  111°  42',        A=    77°  4.6',        a=    95°  50'. 

9.  Impossible. 

10.    0  =  146°  37.9',     B=    55°  0.6',        6=    96°  34.4'. 

Art.  172;   page  133. 

2.  5  =  114°  50',  A=    79°  20',  a=    82°  56'. 

3.  a=    G7°24',  0=164°  6.4',  c  =  160°  6.4' ; 
or,  a  =  112°  36',  0=128°  20.6',  c  =  103°  2.4'. 

4.  c=    90°,  B=    63°  42.7',  b=    66°  26.2'. 

5.  Impossible. 

6.  b=    27°  22.1',  A=  47°  21.2',  a  =117°    9.2'. 

7.  a=    43°  3.1',  B=  89°  23.8',  &  =  129°    8.4'; 
or,  a  =136°  56.9',  B=  26°58.6',  6=    20°35.8'. 

8.  Impossible. 

Art.  173;   pages  134  and  136. 

1.  Distance,  3275  miles;  bearing  of  Greenwich  from 
Boston,  N.  53°  7'  E.  ;  of  Boston  from  Greenwich,  N.  71° 
39.4'  W. 

2.  Distance,  11010  miles  ;  bearing  of  Calcutta  from  Val- 
paraiso, S.  64°  20.4'  E.  ;  of  Valparaiso  from  Calcutta,  S.  54° 
54.6'  W. 

3.  Latitude  49°  58.4' N. 

4.  6  h.,  0  m.,  40  s.,  a.m.  ;  longitude  44°  50'  W. 

5.  15°  0.8'. 

6.  8  h.,  2  m.,  40  s.,  a.m. 

7.  5  h.,  3  m.,  26.4  s.,  a.m. 


1 

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